**For reference, my motivation:** It's of interest to classify free actions of groups on spheres of positive even dimension. Establishing such a classification up to homotopy is not too difficult: Every free group action on a sphere of even dimension is homotopic to either the trivial action by the trivial group or the antipodal action by $\mathbb{Z}/2$. The question in the title is, by the reduction that follows, equivalent to whether this classification is conserved if the "homotopic" in the above sentence is strengthened to "homeomorphic".

Suppose that every space doubly covered by $S^{2n}$ has the homeomorphism type of $\mathbb{P}^{2n}_{\mathbb{R}}$ and let $\tau\ \colon S^{2n}\to S^{2n}$ be some continuous involution lacking fixed points. Then:

By the compactness of $S^{2n}$, $$x\mapsto\text{dist}_{\text{standard subspace Euclidean metric on }S^{2n}}\left(x,\tau\left(x\right)\right)\colon S^{2n}\to\mathbb{R}_{\geq 0}$$ attains a nonzero mminimum on its domain.

By (1), the projection map $$\gamma\ \colon S^{2n}\to\text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)$$ is a covering map.

By (2), there exists an isomorphism $$\psi\ \colon \text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)\to\mathbb{P}_{\mathbb{R}}^{2n}.$$

By (2) and the lifting theorem for covering spaces, the $\psi$ of (3) lifts to an isomorphism $$\tilde{\psi}\ \colon S^{2n}\to S^{2n}$$ such that $$\text{anti}\circ\tau = \tau\circ \tilde{\psi}$$ (with $\text{anti}\ \colon S^{2n}\to S^{2n}$ the antipodal involution), precisely the desideratum.

ycovered"? $\endgroup$nontriviallydoubly covered by $S^{2n}$"? The double cover can't be trivial since $S^{2n}$ is connected. $\endgroup$