For any group $G$, cohomology can be viewed as a functor
$$
H^\ast(G,-): G{\sf\text{-}mod}\to {\sf GrAbGrp},
$$
where $G{\sf\text{-}mod}$ denotes the category of (left) $\mathbb{Z}[G]$-modules and ${\sf GrAbGrp}$ denotes the category of (non-negatively) graded abelian groups.


It seems that there are non-isomorphic groups $G\_1$ and $G\_2$ whose integral group rings $\mathbb{Z}[G\_1]$ and $\mathbb{Z}[G\_2]$ are Morita equivalent (this is a result of [Roggenkamp and Zimmermann](http://www.sciencedirect.com/science/article/pii/002240499500121C#)). One could then ask the following question:



> 
> Do there exist non-isomorphic groups $G\_1$ and $G\_2$ for which there is an equivalence of categories $F: G\_1{\sf\text{-}mod}\to G\_2{\sf\text{-}mod}$ such that the functors $H^\ast(G\_1,-)$ and $H^\ast(G\_2,-)\circ F$ from $G\_1{\sf\text{-}mod}$ to ${\sf GrAbGrp}$ are naturally isomorphic?
> 
> 
> 


This is my (possibly naive) attempt to formalise the question in the title. It may be that the answer is trivially "no" by looking at $H^0$, ie the functor of coinvariants. In that case, I would like to know if there are other formulations for which the question becomes interesting. 



Do there exist non-isomorphic groups with the same cohomology?
The answer would seem to be yes if I understood your question properly. The paper <http://www.m-hikari.com/ija/ija-password-2009/ija-password9-12-2009/ladraIJA9-12-2009.pdf> shows that if $\mathbb ZG$ is isomorphic to $\mathbb ZH$, then one can choose an isomorphism which is augmentation preserving. They then show that if $f\colon \mathbb ZG\to \mathbb ZH$ is the augmentation-preserving isomorphism, then viewing a $\mathbb ZH$-module as a $\mathbb ZG$-module gives isomorphisms in homology and cohomology. I didn't check if their proof gives naturality of the isomorphisms, but I would be surprised if it didn't. 


The paper <http://www.jstor.org/stable/3062112> shows there are nonisomorphic finite groups with isomorphic integral group rings.


