For which $n$ can one infer from these assumptions that there is a $y \in S^n$ such that $f(y) = y$? In this problem $S^n$ means $\{x \in \mathbb{R}^{n + 1} \ | \ \|x\| = 1\}$ and $f\colon S^n \rightarrow S^n$ is a continuous map that satisfies $f(x) = f(-x)$ for every $x \in S^n$.  For which $n$ can one infer from these assumptions that there is a $y \in S^n$ such that $f(y) = y$?
 A: If $f$ is smooth, the answer is "every n".
Here's the key lemma, which works for continuous maps:

Suppose $f:S^n\rightarrow S^n$ is a continuous function with no fixed points.  Then $f$ is homotopic to the antipodal map.  In particular,  the degree of $f$ is $(-1)^{n+1}$.

(The proof of the key lemma is in Hatcher, or see http://www.math.toronto.edu/mat1300/degree.14.pdf Proposition 16.14).
Now, suppose $f$ is a smooth function satisfying $f(-x) = f(x)$.  By Sard's theorem, there is a regular value $p$ of $f$.  Regularness implies that $d_x f$ is linear isomorphism for each $x\in f^{-1}(p)$, and therefore, that $f$ is a local diffeomorphism near each preimage of $p$.  In particular, $f^{-1}(p)$ is a finite.
But, it's not just finite set, but one of even cardinality.  For, if $x\in f^{-1}(p)$, then so is $-x$.
Then, the degree of $f$ is sum over preimages of $p$ of $\pm 1$, depending on whether $f$ preserves or reverses orientation near $x$.  But a sum of an even number of $\pm 1$ is congruent to $0$ mod $2$, so is not equal to $(-1)^n$.  Hence, by the key lemma, $f$ has a fixed point.
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If $f$ is not smooth, I'm not sure what to do.  If you can approximate $f$ by a smooth map while still satisfying the $f(-x) = f(x)$ condition, the above proof works.  Alternatively, I suspect the condition $f(-x) = f(x)$ already implies the degree of $f$ is even, even in the continuous category.  I just don't see how to prove it.
