If $G$ is acting freely on $\mathbb S^{2n}$ we can associate to $\rho\in G, \rho\neq0$ it's degree $deg\rho=(-1)^{n+1}$ as any map without fixed points has that degree. This shows that every non-trivial element is homotopic to the antipodal map. Why does that imply that $G\subset \mathbb Z_2$?
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2The degree should be $(-1)^{2n+1}=-1$. So look at the composition $G\rightarrow\operatorname{Homeo}(X)\rightarrow{\pm1}$ (first the action, then the degree). This is a group homomorphism, what can you say about its kernel? – Thorgott Jan 19 '21 at 13:47
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Recall that $\{-1, 1\}$ is a group with respect to the standard number multiplication, isomorphic to $\mathbb{Z}_2$.
With that we have the following group homomorphism
$$f:G\to\{-1, 1\}$$ $$f(g)=deg(x\mapsto gx)$$
What you've shown is that $f(g)=-1$ for any $g\neq e$. In particular $\ker f=\{e\}$ and thus $f$ is injective.
freakish
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