**Setting**

Let $p: E \rightarrow B$ be a flat fiber bundle with fiber $F$ where $E$, $B$, $F$ are compact, smooth manifolds. Then $E$ has the form of a twisted product

$E \cong \widetilde{B} \times_{\pi_{1}} F$,

where $\widetilde{B}$ is the universal cover of $B$, $\pi_{1}$ is the fundamental group of $B$ and $F$ carries a $\pi_{1}$-action.

We consider cohomology with field coefficients.

**Background**

$\pi_{1}$ finite would be very convenient, for instance then we'd have $$H^*(E) = H^*(\widetilde{B} \times F)^{\pi_{1}} = (H^*(\widetilde{B}) \times H^*(F))^{\pi_{1}}. $$ (Since then $\widetilde{B} \times F$ is compact and the diagonal action is free.) This product structure on the cohomology of $E$ can be very useful.

Sadly, $\pi_{1}$ is often infinite. However, sometimes we can still reduce the structure group to a finite one. If, for instance, there is a triangulation on $F$ such that the $\pi_{1}$-action on $F$ is simplicial, we can do the following:

The group action is in this case a group homomorphism $$ \phi: \pi_1 \rightarrow \{\text{Permutations of Vertices} \}, $$ 
where the right hand side is finite because $F$ is compact. In particular $G:= \pi_1/ker(\phi) $ is finite. We have $$ E \cong (\widetilde{B}/ker(\phi)) \times_{G} F $$
and thus reduced our structure group to a finite one. The diagonal action is still free and $\widetilde{B}/ker(\phi) \times F$ is compact, so we get $$H^*(E) = H^*(\widetilde{B}/ker(\phi) \times F)^{G} = (H^*(\widetilde{B}/ker(\phi)) \times H^*(F))^{G}. $$

**Question**

Are there compact flat fiber bundles where we cannot do this trick to compute cohomology? I am looking for an explicit example. (Since we want to compute cohomology and perhaps characteristic classes, we are only concerned with fiber bundles up to fiber preserving homotopy equivalences.)

**Related**
In this post https://mathoverflow.net/questions/281140/simplicial-structure-on-a-flat-fiber-bundle I learned, that we can always replace $F$ by a simplicial complex (roughly $F \times E\pi_1$) but loose compactness. (So the above trick does not work with this.)