group actions on spheres 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?

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. 
