TITLE: non-orientability of vector bundles induced from a symmetric group action
QUESTION [6 upvotes]: Let $\Sigma_k$ be the symmetric group on $k$-letters. Let $M$ be a manifold with a free $\Sigma_k$-action. Then we can form a $k$-dimensional vector bundle
$$
\xi:\mathbb{R}^k\longrightarrow M\times_{\Sigma_k}\mathbb{R}^k\longrightarrow M/\Sigma_k.
$$
We notice that the transition functions of $\xi$ are given in the form
$$
(a_{i,j}=\delta_{j,\sigma(i)})_{k\times k}
$$
where $\sigma\in \Sigma_k$ and $\delta_{j,\sigma(i)}=1$ if $j=\sigma(i)$ and $0$ otherwise. These matrices have determinants both $1$ and $-1$. 
Question. For any $k\geq 2$, can we conclude that $\xi$ is always non-orientable? 

Thanks for the answer given by Will Sawin! In his solution, I do not understand the following part:
Suppose $M$ is connected and the covering map from $M$ to $M/\Sigma_k$ induces a  surjective homomorphism 
\begin{eqnarray*}
h: \pi_1(M/\Sigma_k)\longrightarrow \Sigma_k. 
\end{eqnarray*}
(I obtained this surjective homomorphism by Prop. 1.40 (c), Algebraic Topology, A. Hatcher). 
Let $r: \Sigma_k\longrightarrow O(k)$ be the regular representation of $\Sigma_k$ given by permuting the coordinates of $\mathbb{R}^k$.  
Question: Does a $k$-dimensional vector bundle over $M/\Sigma_k$ with structure group $\Sigma_k$ is uniquely determined by a map from $\pi_1(M/\Sigma_k)$ to $\Sigma_k$?  Why the bundle $\xi $ comes from the map 
 $r\circ h$?

REPLY [2 votes]: Yes if $M$ is connected, no otherwise.
If $M$ is connected, the covering gives a surjective homomorphism $\pi_1(M/\Sigma_k)\to \Sigma_k$. The vector bundle comes from a representation of $\pi_1$ defined by composing this with the permutation representation of $\Sigma_k$. Taking determinants, the determinant line bundle is the same homomorphism composed with the sign representation of $\Sigma_k$. Because the homomorphism is surjective and the sign representation is nontrivial, the determinant line bundle is nontrivial. (Here I am using the fact that the map from homomorphisms $\pi_1(X) \to \pm 1$ to isomorphism classes of line bundles is a bijection.)
If $M$ is disconnected then the vector bundle could be trivial. For instance, $M$ might itself equal the symmetric group.