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This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$ EG(Y) \times_{G(Y)} Y \to BG(Y) $$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop the total singular complex of $BG(Y)$.

3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$ Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) $$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$ EG\times_G Z \to BG $$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$ EG(Y) \times_{G(Y)} Y \to BG(Y) $$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$.

3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$ Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) $$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$ EG\times_G Z \to BG $$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

The following might lead to an answer of your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...

Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map: $$ B \to B\text{aut}(Y) \to BG(Y) \simeq BG(X) . $$ So if we consider the universal (quasi-)fibration $$ F \to EG(X) \times_{G(X)} F \to BG(X) $$ with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.

Note added: the result I attributed to Peter May is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).

They show if $n$ is large that $$ [B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)] $$ is surjective, where $B$ is finite dimensional CW and $n$ depends on $\dim X$ and $\dim B$. This shows that any finite skeleton of $BG(X)$ is a retract of $B\text{diff}(M \times \Bbb R^j)$ for suitable choice of $j$.

Since $B\text{diff}(M) \to BG(M)$ factors through $B\text{homeo}(M)$, we get the same statement for homeomorphisms. Finally, we can take $j \to \infty$. This will show that $BG(X)$ is a retract of $B\text{homeo}^{\text{st}}(M)$, where $$ \text{homeo}^{\text{st}}(M) = \lim_j \quad \text{homeo}(M\times \Bbb R^j) . $$ By the way, this last group maps to $\text{homeo}(M \times Q)$ where $Q$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.

Yet another note: A possibly related result due to Chapman: If $X$ is $1$-connected CW, then the map $$ \text{homeo}(X\times Q) \to G(X \times Q) \simeq G(X) $$ is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that $f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.

This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$ EG(Y) \times_{G(Y)} Y \to BG(Y) $$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop the total singular complex of $BG(Y)$.

3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$ Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) $$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$ EG\times_G Z \to BG $$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

The following might lead to an answer of your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...

Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map: $$ B \to B\text{aut}(Y) \to BG(Y) \simeq BG(X) . $$ So if we consider the universal (quasi-)fibration $$ F \to EG(X) \times_{G(X)} F \to BG(X) $$ with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.

Note added: the result of Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).

They show if $n$ is large that $$ [B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)] $$ is surjective, where $B$ is finite dimensional CW and $n$ depends on $\dim X$ and $\dim B$. This shows that any finite skeleton of $BG(X)$ is a retract of $B\text{diff}(M \times \Bbb R^j)$ for suitable choice of $j$.

Since $B\text{diff}(M) \to BG(M)$ factors through $B\text{homeo}(M)$, we get the same statement for homeomorphisms. Finally, we can take $j \to \infty$. This will show that $BG(X)$ is a retract of $B\text{homeo}^{\text{st}}(M)$, where $$ \text{homeo}^{\text{st}}(M) = \lim_j \quad \text{homeo}(M\times \Bbb R^j) . $$ By the way, this last group maps to $\text{homeo}(M \times Q)$ where $Q$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.

Yet another note: A possibly related result due to Chapman: If $X$ is $1$-connected CW, then the map $$ \text{homeo}(X\times Q) \to G(X \times Q) \simeq G(X) $$ is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that $f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.

The following might lead to an answer of your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...

Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map: $$ B \to B\text{aut}(Y) \to BG(Y) \simeq BG(X) . $$ So if we consider the universal (quasi-)fibration $$ F \to EG(X) \times_{G(X)} F \to BG(X) $$ with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.

Note added: the result I attributed to Peter May is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).

They show if $n$ is large that $$ [B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)] $$ is surjective, where $B$ is finite dimensional CW and $n$ depends on $\dim X$ and $\dim B$. This shows that any finite skeleton of $BG(X)$ is a retract of $B\text{diff}(M \times \Bbb R^j)$ for suitable choice of $j$.

Since $B\text{diff}(M) \to BG(M)$ factors through $B\text{homeo}(M)$, we get the same statement for homeomorphisms. Finally, we can take $j \to \infty$. This will show that $BG(X)$ is a retract of $B\text{homeo}^{\text{st}}(M)$, where $$ \text{homeo}^{\text{st}}(M) = \lim_j \quad \text{homeo}(M\times \Bbb R^j) . $$ By the way, this last group maps to $\text{homeo}(M \times Q)$ where $Q$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.

Yet another note: A possibly related result due to Chapman: If $X$ is $1$-connected CW, then the map $$ \text{homeo}(X\times Q) \to G(X \times Q) \simeq G(X) $$ is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that $f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.