Existence of a map in a Hilbert space Let $H$ be an infinite-dimensional Hilbert space, $B$ be its unit ball: $B=\{x\in H: \, \|x\|\leq 1\}$.
Does there exist a continuous map $f:H\to H$ such that $f(f(x))=x$ $\forall x\in H$, $f$ has no fixed points,
and $f(B)$ is unbounded?
 A: Such involutions $f$ do exist. I will explain the construction without the part dealing with $f(B)$ (this restriction does not appear to be very natural); if there is enough interest I can add more details concerning $f(B)$. First of all, every infinite dimensional Hilbert space $H$ is diffeomorphic to its unit sphere $S$, see the discussion and references to Bessaga's theorem in this MSE post. Let $g: S\to H$ be such a diffeomorphism. Now consider the involution $i: S\to S$, $i(x)=-x$. It is clear that $i$ has no fixed points. Lastly, take the composition $f= g \circ i \circ g^{-1}$. This is a self-diffeomorphism of $H$ which is involutive and has no fixed points. 
As for the finite-dimensional case: 
Every prime order periodic self-homeomorphism of a finite-dimensional contractible CW complex $X$ has a fixed point.
For, otherwise,  the group $G=Z/p$ ($p$ is prime) would have a finite-dimensional $K(G,1)$, namely, $X/G$, which would imply that $G$ has finite cohomological dimension. But all finite nontrivial groups have infinite cohomological dimension (this is explained, for instance, in Brown's book "Cohomology of groups"). Now apply this result to $X=R^n$ and $p=2$. 
