A proof is on page 80-81 and 90 of Forrester, the probability distribution function of $W=X^\top X$ is $\propto e^{-\tfrac{1}{2}\operatorname{tr}W}(\operatorname{det}W)^{(n-m-1)/2}$, for an $n\times m$ matrix $X$, so only dependent on the eigenvalues of $W$. The $m\times m$ matrix of eigenvectors of $W$ is thus uniformly distributed in the orthogonal group, independently of the eigenvalues.

A proof is on page 80-81 and 90 of Forrester, the probability distribution function of $W=X^\top X$ is $\propto e^{-\tfrac{1}{2}\operatorname{tr}W}(\operatorname{det}W)^{(n-m-1)/2}$, for an $n\times m$ matrix $X$, so only dependent on the eigenvalues of $W$. The $m\times m$ matrix of eigenvectors of $W$ is thus uniformly distributed in the orthogonal group, independently of the eigenvalues.

You can also see this directly from the fact that $P(X)\propto e^{-\tfrac{1}{2}\operatorname{tr}X^\top X}$, so invariant under $X\mapsto XO$ with orthogonal $O$.