TITLE: Constructing a odd homeomorphism between $A$ and $S^n$
QUESTION [6 upvotes]: I have asked this question here seven months ago and until now I got no answer.
Let $A\subset\mathbb{R}^N\setminus\{0\}$ be a closed symmetric set ($x\in A$ then $-x\in A$). Suppose that $A$ is homeomorphic to some sphere $S^n$, $n\leq N$ ($n$ is the dimension of the sphere). Is it possible to construct a homeomorphism $F:A\to S^n$ such that $F$ is odd?

REPLY [7 votes]: The answer is "NO".
It follows from existence of exotic smooth involutions of sphere. 
Say, this movie explains a construction of an involution  $\iota$ of $\mathbb S^5$ such that
the quotient $\Pi=\mathbb S^5/\iota$ is homotopy equivalent, but not not homeomorphic to $\mathbb R\mathrm P^5$.
Consider composition
$$\Pi\longrightarrow^{\!\!\!\!\!h}\mathbb R\mathrm P^5\longrightarrow^{\!\!\!\!\!e}\mathbb R\mathrm P^{N-1},$$ 
where $h$ is the homotopy equivalence and $e$ is the standard embedding.
If $N$ is large, Whitney embedding theorem says that you can perterb this composition into smooth embedding $\Pi\to\mathbb R\mathrm P^{N-1}$. 
Passing to the double cover of $\Pi$ and $\mathrm P^{N-1}$,
you get a funny embedding $\mathbb S^5\hookrightarrow \mathbb S^{N-1}\subset \mathbb R^N$, 
which gives a counterexample.
(If the image admits is an odd homeomorphism to $\mathbb S^5$ then 
$\Pi$ has to be homeomorphic to $\mathbb R\mathrm P^5$.)