Let $X$ be a manifold acted on by a Lie group $G$. The $G$-equivariant cohomology of $X$ with coefficients in a ring $\mathcal{R}$ is defined as the cohomology ring
$$
H_G^*(X; \mathcal{R}) := H^*(X_G; \mathcal{R}),
$$
where $X_G := (X \times EG) / G$ is the **homotopy quotient**, $EG \to BG = EG / G$ being the universal principal bundle of the group $G$.

The natural projection $$ X_G \to BG $$ obtained by collapsing the elements of $X$ gives rise to a ring homomorphism $$ H^*(BG; \mathcal{R}) \to H_G^*(X; \mathcal{R}), $$ or equivalently to an action of the ring $H^*(BG; \mathcal{R})$ on the equivariant cohomology $H_G^*(X; \mathcal{R})$ of $X$.

Suppose now that $G$ acts freely on $X$. In this case, the cohomology groups $H_G^*(X; \mathcal{R})$ and the singular cohomology $H^*(X/G; \mathcal{R})$ of $X/G$ agree.

My question is the following: how does $H^*(BG; \mathcal{R})$ act on the singular cohomology of $X/G$ (if it can make things easier, one might take $\mathcal{R}=\mathbb{C}$)?