Are the properties of not having fixed points and being its own inverse enough to characterise the antipodal map?

Question

While trying to solve a problem I ended up with a smooth map $f:S^n \to S^n$ which I know is fixed point free and such that $f \circ f $ is the identity map of the sphere. I would like to conclude this map is the antipodal map of $S^n$, but while I can't find a reasonable argument to prove this I can't think of any other functions with these two properties. Is it true that the only map satisfying these conditions is the antipodal map of $S^n$?

Answer 1

Up to homotopy, yes. It's well known that a fixed-point free map from $S^n$ to $S^n$ is homotopic to the antipodal map via $$ F_t(x) = \frac{(1-t)f(x) -tx}{\|(1-t)f(x)-tx\|}.$$

Up to equality, no. In Some curious involutions of spheres, which is available here, Hirsch and Milnor considered the question of whether the quotient of a sphere by a fixed-point free involution was isomorphic to projective space (which would be the case if the involution were the antipodal map).

The first two lemmas in that paper construct differentiable, fixed-point free involutions of $S^6$ and $S^5$ whose quotients are not diffeomorphic to projective space. Thus the answer is no.

I should note that the authors also mention that it is true that for $n\le 3$ the answer to their question is positive, which likely explains the difficulty of constructing such involutions.