There is a precise relation at the level of complexes: $C^\ast(X,\mathbb Z)$ is a $G$-complex and as such it is perfect (that is quasi-isomorphic to a finite complex consisting of projective modules) and furthermore $C^\ast(X/G,\mathbb Z)$ is quasi-isomorphic to the derived functor value $R\Gamma(G,C^\ast(X,\mathbb Z)$. The latter is mostly used through its consequent spectral sequence $H^\ast(G,H^\ast(X,\mathbb Z))\implies H^\ast(X/G,\mathbb Z)$ but (as with all spectral sequences) it contains some ambiguity which (somehow) has to be resolved.

This is mainly going from knowledge of the cohomology of $X$ to that of $X/G$. Going the other direction is more difficult as the $G$-cohomology kills a lot of information (if one works with rational coefficients instead, it just picks out the trivial representations). However, the fact that $C^\ast(X,\mathbb Z)$ is perfect helps out even though it can still be difficult to say something.

As an example of the relevance of perfectness consider the case when $G$ is cyclic (of order $n$, say) acting by fixed point free orientation preserving maps on the $k$-sphere. Then the cohomology of $X$ is the trivial representation in degrees $0$ and $k$. Such a complex is classified (this is essentially the Yoneda Ext-description) by an element $\alpha$ in $H^{k+1}(G,\mathbb Z)$. For $C^\ast(X,\mathbb Z)$ to be perfect we must have that $\alpha$ must have order exactly $n$. This excludes $k$ even as the order then is always $1$ and for $k$ odd $\alpha$ must be a generator of $H^{k+1}(G,\mathbb Z)=\mathbb Z/n$. It is now easy to compute $R\Gamma(G,C^\ast(X,\mathbb Z)$ (and its additive structure is actually independent of $\alpha$) but we have also obtained a (necessarily) non-trivial invariant of the action. When $k=3$ this is a well-known invariant of lens spaces.

`$H^*(X/G)$`

is the $G$ invariants of`$H^*(X)$`

). Example: take the covering space`$S^2\to\mathbb{R}P^2$`

with integer coefficients. Then $H^2(X)=0$ but $H^2(X/G)\neq0$. – David Sprehn Mar 2 '11 at 4:44`$H^*(G)$`

to be the singular cohomology of $G$, and others the group cohomology. I don't know which one the questioner intended (but group cohomology would make more sense). – David Sprehn Mar 2 '11 at 5:14