Let $f: E\rightarrow B$ be a Kan fibration between pointed connected Kan complexes with fibre the Eilenberg-MacLane space $\mathrm{K}(M, n), n\geq 2, M$ an abelian group. Assume $f$ induces an isomorphism on $\pi\_1$ with common value $G$, and so $G$ acts naturally on $\pi\_n\mathrm{K}(M, n)=M$ hence on (the simplicial abelian group) $\mathrm{K}(M, n)$. Let $\tilde{B}$ be the universal cover of $B$, on which $\pi\_1B=G$ acts.



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> Is there a fibration equivalence $\tilde{B}\times\_G\mathrm{K}(M, n)\sim E$ over $B$?
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About fibrations with fibre Eilenberg-MacLane spaces
No. If this were the case then there would be a section $s: B \to E$ to $f$ induced by the $G$-equivariant map $\widetilde{s}:\widetilde{B} \to \widetilde{B} \times {\rm K}(M,n)$ sending $x$ to $(x,0)$ (where $0 \in {\rm K}(M,n)$ is the neutral element, which is fixed by the action of $G$). However, there exists fibrations with fibers ${\rm K}(M,n)$ for $n \geq 2$ which do not admit a section, such as the quaternionic Hopf fibration $S^7 \to S^4$. In general fibrations $E \to B$ with fiber ${\rm K}(M,n)$ with a fixed $\pi\_1(B)$-action on $M$ are classified by the cohomology group with local coefficients $H^{n+1}(B,M)$, where the ones of the form you describe correspond to the $0$ element.


**Edit**:


As pointed in the remarks below, the quaternionic Hopf fibration is indeed not a counter-example for what the OP is asking since its fibers are $S^3$, which is not an Eilenberg-MacLane space. One way to fix this is to take the relative Postnikov truncation (which still doesn't have a section). Alternatively, there are probably other more natural examples.


