Let $\mathbb S^n$ be the $n$-sphere: $$\mathbb S^n=\left\{x \in \mathbb R^{n+1}: \left\|x\right\|=1\right\}.$$The **hairy ball theorem** can be formulated as follows:

If $n$ is even and $f\,\colon\, \mathbb S^n \to \mathbb S^n$ is a continuous function, then there exists at least one $x \in \mathbb S^n$ such that either $f(x)=x$ or $f(x)=-x$.

This is not true for odd $n=2k-1$, with a counterexample being $$f(x_1,\,x_2,\,\dots,\,x_{2k-1},\,x_{2k})=(-x_2,\,x_1,\,\dots,\,-x_{2k},\,x_{2k-1}).$$

But what if remove the evenness condition for $n$ and demand $f$ to be even instead? Is the following statement true?

Let $n \in \mathbb N_0$ and $f\,\colon\, \mathbb S^n \to \mathbb S^n$ be a continuous function such that $f(x)=f(-x) \;\; \forall x \in \mathbb S^n$. Then $f$ has a fixed point.

This, of course, is true for even $n$-s, being a particular case of the hairy ball theorem. It is not hard to prove it also for $n=1$, but what about larger odd $n$-s?