vector bundles induced by an action of a finite subgroup of $O(n)$

Question

Let $M$ be a path-connected manifold. Let $G$ be a finite subgroup in $O(n)$ and suppose $G$ acts freely on $M$. Then we have an associated vector bundle $$ \xi(M,G): \mathbb{R}^n\longrightarrow M\times_{G} \mathbb{R}^n\longrightarrow M/G. $$ And by the covering space theory, we have an epimorphism $$ r(M,G): \pi_1(M/G)\longrightarrow G. $$

Question: Let the group $G$ be fixed. Suppose we have manifolds $M_1,M_2$ and bundles $\xi(M_1,G), \xi(M_2,G)$ satisfying the above conditions. If $$ M_1/G\cong M_2/G $$ as spaces and $$ r(M_1,G)=r(M_2,G) $$ as epimorphisms, can we conclude that $\xi(M_1,G)\cong \xi(M_2,G)$ as vector bundles?

Answer 1

By the universal property of the universal covering $U$ of $M_1/G$ you have epimorphisms from $U$ to both $M_1$ and $M_2$. The groupp of deck transformations $D$ of the covering $U\to M_1$ is the kernel of $r(M_1,G)$ and likewise for $M_2$, so the two kernels are equal, hence $M_1\cong U/D\cong M_2$ and the vector bundle in question is the one induced from the representation $r_1$ of the fundamental group of $M_1/G$. So, yes, the two vector bundles are isomorphic.