group actions on spheres

Question

Let $\mathbb{Z}/2$ act on the $m$-sphere $S^m$ freely and properly discontinuously. If the action is not trivial, can we conclude that the action is homotopy equivalent to the antipodal action? That is, the following diagram

commutes up to homotopy?

Can we generalize this to $S^1$-actions on $S^{2m+1}$ and $S^3$-actions on $S^{4m+3}$? Is the uniqueness true?

Answer 1

If $\Bbb Z/2$ acts freely on $S^n$, then acting by the nontrivial element of $\Bbb Z/2$ one gets a map $f:S^n \to S^n$ which has no fixed point.

Thus, $f(x) \neq x$ for all $x \in S^n$. The homotopy $$F(x, t) = \frac{(1-t)x - tf(x)}{|(1-t)x - tf(x)|}$$

between $f$ and the antipodal map $-\text{id}$ is well-defined. Thus the map induced by that action is indeed homotopic to the antipodal map.

Answer 2

No you cannot. For example your action could consist of a rotation by 180 degrees around some hyperplane. That is homotopic to the identity but not the antipodal map.