The following is a question from a qualifying exam I am studying for:
Let $G$ be a group of homeomorphisms acting freely on $S^{2n}$. Show if $G$ has order $2$, then the nontrivial element must be the antipodal map.
Here is what I have so far. If $f$ is the nontrivial element, we know that $f\simeq \alpha$, and $f$ has degreee $(-1)^{2n+1}=-1$, and that $f$ and $\alpha$ must agree on some point of $S^{2n}$. I am skeptical that degree theory will be enough to tackle this problem.
I'm not convinced this is true. Let's try and construct a homeomorphism of $S^2$ which is fixed-point-free and squares to the identity.
Let $\alpha$ be the antipodal map, and let $\phi$ be a homeomorphism of $S^2$ which is the identity away from some small disk $U$ about the north pole, but which is not the identity on $U$. Now consider $f:=\alpha \phi^{-1} \alpha \phi \alpha$ (intuitively, it's the antipodal map but we've modified it a bit on $U$ and made a corresponding change on $\alpha (U)$). This is certainly a homeomorphism, and squares to the identity since $f^2 = \alpha \phi^{-1} \alpha \phi \alpha^2 \phi^{-1} \alpha \phi \alpha$, which all cancels away. Moreover it's fixed-point-free since it swaps the northern and southern hemispheres, so any potential fixed point must lie on the equator, but it's the antipodal map on the equator.