= Solution
We first prove that <fixed-point-free sphere maps are homotopic to the antipodal map>. If $f:S^d\to S^d$ has no fixed point, define
$$
H_t(x)=\frac{(1-t)f(x)-tx}{\|(1-t)f(x)-tx\|},
\qquad 0\leq t\leq1.
$$
The denominator could vanish only if $f(x)$ is a positive multiple of $x$. Both are unit vectors, so this would require $f(x)=x$ and $t=1/2$, excluded by the hypothesis. Thus $H$ is a <continuous> <homotopy> from $f$ to $x\mapsto-x$.
On $S^{2r}$, with $r\geq1$, the antipodal map has degree $(-1)^{2r+1}=-1$, since it is the boundary map of the ambient linear transformation $-I$ on $\mathbb R^{2r+1}$. Homotopy invariance therefore gives degree minus one for every fixed-point-free action map. The degree of each <homeomorphism> is $\pm1$, and functoriality makes degree a <group homomorphism>
$$
G\longrightarrow\{1,-1\}.
$$
Every nonidentity element in a free action has degree minus one, so this homomorphism has trivial kernel. Hence
$$
\boxed{|G|\leq2.}
$$
For $S^0$ the same bound follows immediately from its two points. This proves that <free actions on even-dimensional spheres have order at most two>, without an Euler-characteristic argument requiring a cell structure on the quotient.
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