Given a space $X$ and an action of a group $G$ on $X$, the $G$-invariant cochains with coefficients in an Abelian group $A$ define a sub-cocomplex $\mathcal{C}^{\bullet}_G$ of the cocomplex $\mathcal{C}^{\bullet}$ of cochains with coefficients in $A$. I will call the cohomology $H^{\bullet}(\mathcal{C}_G)$ the "invariant cohomology" of $X$. What is the relationship between the follwing three objects: (i) the invariant cohomology, (ii) the usual equivariant cohomology $\mathcal{H}_G^{\bullet}(X, A)$, and (iii) the cohomology of the quotient space $H^{\bullet}(X/G,A)$? Note that I am most interested in the case where $A$ is a finite Abelian group.

I think that there are always homomorphisms $$H^n(X/G,A) \to H^n(\mathcal{C}_G) \to \mathcal{H}_G^n(X,A).$$ The first map comes from pulling back the projection $X \to X/G$, and the second can be seen from the interpretation of equivariant cohomology as the total cohomology of the double cocomplex of group cochains of $G$ valued in $\mathcal{C}^{\bullet}$. But how does one characterize the kernel and image of these maps? Is there a simple statement if $X$ is contractible, say?

[Edit: regarding the relationship between $H^n(X/G,A)$ and $H^n(\mathcal{C}_G)$, please see the comments on Mark Grant's answer. The upshot is that with simplicial cochains, $H^n(\mathcal{C}_G)$ seems to depend on the choice of triangulation. But, if one defines $\mathcal{C}_G$ in a suitably triangulation-independent way, i.e. either singular cochains, or simplicial cochains on the abstract simplicial complex containing *all* simplex embeddings into X, then $H^n(\mathcal{C}_G)$ and $H^n(X/G,A)$ are almost certainly not the same.]