Is there only one free (continuous) action of $\mathbb{Z}_2$ on $S^2$? We all know that the antipodal map is a free action of $\mathbb{Z}_2$ on $S^2$. Considering $\mathbb{Z}_2 = \{1, -1\}$, a free action may be viewed as a map $f : S^2 \rightarrow S^2$, i.e. the action of $-1$, with no fixed points, such that $f \circ f = Id_{S^2}$. We then readily see that $deg(f) = \pm 1$. Since $f$ does not have fixed points, it cannot have degree 1, and hence $f$ is homotopic to the antipodal map.
My question is: does this imply that the quotient of $S^2$ by this action is homeomorphic to $\mathbb{RP}^2$? Actually, I know this is true, because the quotient must be a closed surface with fundamental group $\mathbb{Z}_2$, so we can use the classification of closed surfaces. But I would like to be able to generalize this to spheres of any even dimension, so I wanted to see if I could get an answer that doesn't use that.
 A: Unfortunately, the result is not true in dimensions of the form $4k$.  That is, in each dimension of the form $4k$, there is a $Z_2$ action on $S^{4k}$ with quotient not homeomorphic to $\mathbb{R}P^{4k}$.  (I don't know what happens in other dimensions.  I suspect that Ricci flow techniques in dimension $3$ would show that the the only $Z_2$ quotients of $S^3$ is the standard $\mathbb{R}P^3$.  I'm far from an expert in this area, though.)
More specifically, in 

D. Ruberman. Invariant Knots of free involutions of S4, Top. Appl. 18 (1984), 217-224

Ruberman shows that there is a non-smoothable $4$-manifold $X$ which is homotopy equivalent, but not homeomorphic to $\mathbb{R}P^4$.
The universal cover $\tilde{X}$ of $X$ must therefore be homotopy equivalent to $S^4$.  By Freedman's classification, we know $\tilde{X}$ is homeomorphic to $S^4$.  Now, the deck group (which is isomorphic to $\pi_1(X)\cong \pi_1(\mathbb{R}P^2)\cong \mathbb{Z}_2$) acts freely on $\tilde{X}\cong S^4$.
Further, in 

R. Fintushel and R. J. Stern, Smooth free involutions on homotopy 4k-spheres, Michigan Math. J. 30 (1983), no.1, 37–51,

Fintushel and Stern find examples of smooth $4k$ manifolds (with $k>1$) which are homotopy equivalent, but not homeomorphic to $\mathbb{R}P^{4k}$.  One can repeat the argument above (using Smale's proof of the Poincare conjecture in high dimensions instead of Freedman's result) to find the involutions on $S^{4k}$.
A: The quotients of such actions are called "Fake real projective spaces" and you find their classification in http://www.map.mpim-bonn.mpg.de/Fake_real_projective_spaces
