Sheaves with isomorphic cohomology, but not quasi-isomorphic

Question

Suppose I have two (constructible) sheaves of vector spaces $F$ and $G$ over the same base space that have isomorphic cohomology (degree by degree), but no sheaf map inducing this isomorphism (i.e. they are not quasi-isomorphic).

Now imagine I apply any of Grothendieck's 6 operations (or other functors) to $F$ and $G$, will the resulting sheaves, say $\Psi(F)$ and $\Psi(G)$, have isomorphic cohomology as well?

Thanks!

EDIT: I suppose the answer in general is no. Consider an injective (acyclic) sheaf and any functor that doesn't preserve injectives. A concrete example would be nice though.

EDIT 2: Thanks to Algori's comments I need to substantially limit the sheaves of interest. Assume additionally that the sheaves have identical support and furthermore that when we take the function that assigns to a point the euler characteristic of the stalk over that point (thereby producing a constructible function) to these two sheaves, these functions are identical. If we are using real sub-analytic partitions then a theorem of Kashiwara's says that the Grothendieck group of the bounded derived category of real constructible sheaves is isomorphic to the group of constructible functions. If I understand things correctly this would mean that these two sheaves are equivalent in the Grothendieck group (which seems weird since this function ignores gluing data). I am dealing with particular examples of two constructible (cellular) sheaves over a simplicial complex, which are not isomorphic yet produce identical constructible functions.

Answer 1

Then how about this example:

Let $X$ be the wedge of two circles. To give a local system on $X$ with stalk a complex vector space $V$ is to give two matrices $A_1,A_2\in GL(V)$. If the subgroup $\langle A_1,A_2\rangle$ of $GL(V)$ generated by $A_1,A_2$ has no invariants in $V$, then $H^0$ of the resulting local system will be $0$ and $H^1$ will be of rank $\dim V$ for Euler characteristic reasons.

To give a map in $Sh(X)$ between the local systems that correspond to $(V,A_1,A_2)$ and $(W,B_1,B_2)$ is to give a linear map $f:V\to W$ such that $f A_1=B_1f,f A_2=B_2f$. Now suppose that $V=W$, that both the groups $\langle A_1,A_2\rangle$ and $\langle B_1,B_2\rangle$ have no invariants in $V$, that $A_1$ is diagonalizable and that the eigenvalues of $A_1$ and $B_1$ are pairwise distinct. Then any map between the local systems is zero.