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?



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


