Let $\pi$ be a group and $K(\pi,1)$ the EilenbergMacLane space. Let $G$ be a finite group acting on $K(\pi,1)$ such that the following is a covering map $$ K(\pi,1)\longrightarrow K(\pi,1)/G. $$ Question. Is $K(\pi,1)/G$ an EilenbergMacLane space? Does $$ K(\pi,1)/G=K(\pi \times G,1)? $$

Suppose a group $H$, not necessarily finite, acts on an EilenbergMacLane space $BN$. The homotopy quotient $BN/H$ (which agrees with the ordinary quotient if the action of $H$ is free) fits into a fiber sequence $$BN \to BN/H \to BH$$ and the long exact sequence in homotopy shows that $BN/H$ has vanishing higher homotopy. Hence it is an EilenbergMacLane space $BG$ for a group $G$ fitting into a short exact sequence $$1 \to N \to G \to H \to 1$$ (determined by the action). In other words, it's an extension of $H$ by $N$. Every such extension arises in this way. Among them, semidirect products (extensions for which the above sequence splits) correspond to pointed actions of $H$ on $BN$, or equivalently actions of $H$ on $BN$ admitting a (homotopy) fixed point. For example, take $BN$ to be the configuration space of $n$ ordered points in $\mathbb{R}^2$, so that $N = P_n$ is the pure braid group, and $H = S_n$ acts by permutations of the points. Then $BN/H = BG$ is the configuration space of $n$ unordered points in $\mathbb{R}^2$, so that $G = B_n$ is the usual braid group. The corresponding short exact sequence $$1 \to P_n \to B_n \to S_n \to 1$$ does not split. 


$K(\pi,1)/G$ is not necessarily $K(\pi\times G,1)$ for example take $G=Z$, $K(Z,1)=S^1$. $S^1=R/(t_1=x\rightarrow x+1)$ the quotient of $S^1$ by the group $Z/n$ generated by the transformation induced by $x\rightarrow x+1/n$ is $S^1$. 

