= Solution
Assume for contradiction that the <involution> $\phi$ is fixed-point-free. The finite group action is then a <covering space action>, so the quotient map
$$
q:\mathbb R^n\longrightarrow Y=\mathbb R^n/\langle\phi\rangle
$$
is a double covering and $Y$ is an $n$-manifold.
Apply the <long exact sequence in homology> to the <transfer chain map of a double covering>. Since $\mathbb R^n$ is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that $q_*:H_0(\mathbb R^n;\mathbb F_2)\to H_0(Y;\mathbb F_2)$ is an isomorphism, gives
$$
H_1(Y;\mathbb F_2)\cong H_0(Y;\mathbb F_2)\cong\mathbb F_2.
$$
In every higher degree the same exact sequence gives
$$
H_i(Y;\mathbb F_2)\cong H_{i-1}(Y;\mathbb F_2).
$$
Thus $H_i(Y;\mathbb F_2)\cong\mathbb F_2$ for every $i\geq0$. This contradicts <homology above the dimension of a manifold>, which gives $H_i(Y;\mathbb F_2)=0$ for $i>n$. Therefore $\phi$ has a <fixed point>.
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