Let $M$ be a finite $CW$-complex. Let $\Sigma\_k$ be the symmetric group acting on $k$-letters. Suppose there is a free action of $\Sigma\_k$ on $M$. Then we have a covering map 
$$
f:M\to M/\Sigma\_k. 
$$
From $f$, we have an induced map on the fundamental groups
$$
f\_\*:\pi\_1(M)\to \pi\_1(M/\Sigma\_k). 
$$
From $f\_\*$, we have an induced map on the Eilenberg-Maclane spaces
$$
F: K(\pi\_1(M),1)\to K(\pi\_1(M/\Sigma\_k),1).
$$
**Question:** Does there always exist an induced $\Sigma\_k$-action on $K(\pi\_1(M),1)$ such that $F$ is a **covering map** obtained by modulo this $\Sigma\_k$-action on $K(\pi\_1(M),1)$?



induced group actions and covering maps on Eilenberg-Maclane space
If $M$ is connected, then @MarkGrant's fibration sequence gives a long exact sequence on homotopy groups showing that $\pi\_1(M/\Sigma\_k)\to \Sigma\_k$ is surjective. Now apply the $K(-,1)$-functor and obtain a map $K(\pi\_1(M)/\Sigma\_k,1)\to B\Sigma\_k$. The corresponding $\Sigma\_k$-bundle over $K(\pi\_1(M)/\Sigma\_k,1)$ has as total space a $K(\pi\_1(M),1)$, as can be seen again from the long exact sequence in homotopy. 


So there exists a model of your map $F$ which is a covering map. You have defined $F$ only up to homotopy, so I think your question needs to be whether one can find a model of $F$ which is a covering map.


