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What is the normalizer of the group of $C^\infty$ diffeomorphisms on $[0, 1]$, with group law given by composition, in the group of all homeomorphisms of $[0, 1]$?

If the answer is known, is there some "elementary" proof for the result?

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    $\begingroup$ What is the (not commutative) group structure of $C^0[0,1]$? $\endgroup$
    – Dieter Kadelka
    Dec 8 at 12:48
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    $\begingroup$ It refers to only the homeprphisms and smooth diffeomorphisms, respectively, with composition as the group multiplication. $\endgroup$
    – Henry
    Dec 8 at 21:36
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    $\begingroup$ The question is interesting. Maybe you could change the notaion (and title), because usually $C^0[0,1]$ refers to the continuous, real-valued functions on $[0,1]$ and not to the homeomorphisms of $[0,1]$ and similar for $C^\infty$. $\endgroup$
    – tj_
    Dec 9 at 0:03
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    $\begingroup$ I guess it comes from the question of determining the automorphism group of $\mathrm{Diff}^\infty([0,1])$ as Rubin spatiality theorems might reduce to determining this normalizer. I guess it's not trivial and one shouldn't expect a too easy answer. $\endgroup$
    – YCor
    Dec 9 at 9:17
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    $\begingroup$ Filipkiewicz proved in "Isomorphisms between diffeomorphism groups", 1982, that every isomorphism (as abstract groups) between $Diff^k(M)$ and $Diff^k(N)$ is induced by a $C^k$-diffeo between $M$ and $N$, if $1 \leq k \leq \infty$ and $M, N$ are paracompact manifolds without boundary. So, I think, the problem reduces to finding normaliser of $Diff([0, 1])$ inside $Diff((0, 1))$, which should be known. $\endgroup$
    – Denis T
    Dec 9 at 15:56

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