When are these irreducible complex representations for the Type D Weyl group self-dual?

Question

Given an integer $n \geq 4$, consider the Weyl groups $W(B_n)$ and $W(D_n)$ of types $B_n$ and $D_n$, respectively, and consider their representation theory over the field of complex numbers.

The Weyl group $W(B_n)$ is the wreath product $C_2 \wr S_n = C_2^n \rtimes S_n$ of the cyclic group $C_2 = \{\pm 1\}$ and the symmetric group $S_n$. It is well-known that the irreducible $W(B_n)$-modules are parameterized by bi-partitions $(\lambda,\mu)$ of $n$, i.e., ordered pairs of partitions $\lambda$ and $\mu$ such that $|\lambda|+|\mu| = n$. One can show that if $V = V^{(\lambda,\mu)}$ is the irreducible $W(B_n)$-module labeled by the bi-partition $(\lambda,\mu)$, then the dual module $V^* = \text{Hom}_{\mathbb{C}}(V,\mathbb{C})$ is isomorphic to $V^{(\lambda',\mu')}$, where $\lambda'$ and $\mu'$ are the transpose (or conjugate) partitions associated to $\lambda$ and $\mu$, respectively. In particular, $V \cong V^*$ if and only if $\lambda = \lambda'$ and $\mu = \mu'$.

The Weyl group $W(D_n)$ is the kernel of the homomorphism $\xi: W(B_n) = C_2^n \rtimes S_n \to \{\pm 1\}$ that is trivial on $S_n$ and that acts on $\mathbb{Z}_2^n = \{\pm 1\}^n$ by multiplying all of the factors together. In particular, $W(D_n)$ is an index-$2$ subgroup of $W(B_n)$. Since $V^{(\lambda,\mu)} \otimes \xi \cong V^{(\mu,\lambda)}$ as $W(B_n)$-modules, one then gets the following via classical Clifford theory:

1. If $\lambda \neq \mu$, then $V^{(\lambda,\mu)}$ and $V^{(\mu,\lambda)}$ are isomorphic and irreducible as $W(D_n)$-modules.
2. If $(\lambda,\lambda)$ is a bi-partition of $n$, then $V = V^{(\lambda,\lambda)}$ decomposes as a $W(D_n)$-module into a direct sum $V^{(\lambda,\lambda)^+} \oplus V^{(\lambda,\lambda)^-}$ of two non-isomorphic, conjugate, irreducible $W(D_n)$-modules.
3. Every irreducible $W(D_n)$-module is uniquely of one of the preceding forms, subject only to the $W(D_n)$-module isomorphism $V^{(\lambda,\mu)} \cong V^{(\mu,\lambda)}$ if $\lambda \neq \mu$.

I would like to know when an irreducible $W(D_n)$-module is self-dual. The only case where there seems to be some ambiguity is when $n$ is even, for irreducibles of the forms $V^{(\lambda,\lambda)^+}$ and $V^{(\lambda,\lambda)^-}$ with $\lambda = \lambda'$. In this case $V^{(\lambda,\lambda)}$ is self-dual as a $W(B_n)$-module, so it follows that either $V^{(\lambda,\lambda)^+}$ and $V^{(\lambda,\lambda)^-}$ are each self-dual as $W(D_n)$-modules, or one of them is isomorphic to the dual of the other.

Given a symmetric partition $\lambda$, is there a criterion, depending on the shape of the Young diagram of $\lambda$, for deciding whether $V^{(\lambda,\lambda)^+}$ and $V^{(\lambda,\lambda)^-}$ are each self-dual?

An analogous situation occurs when looking at the irrreducible representations of the alternating group $A_n$. If $\lambda = \lambda'$, then the Specht module $S^\lambda$ decomposes into a direct sum $S^{\lambda^+} \oplus S^{\lambda^-}$ of two irreducible $A_n$-modules. These modules are self-dual if and only if their corresponding complex characters are real-valued. There are explicit formulas for the values of these characters (see for example Theorem 2.5.13 in James and Kerber's book The Representation Theory of the Symmetric Group, or Proposition 5.3 in Fulton and Harris' book Representation Theory), from which one can see that the characters will both be real-valued if and only if the number of squares above the diagonal in the Young diagram of $\lambda$ is even.

Answer 1

As noted in comments, all irreducible characters of Weyl groups are rational, and, in particular, real.