Let $G$ be a compact Lie group acting smoothly on two manifolds $M$ and $N$ and suppose we have an equivariant map $f: M \rightarrow N$ which induces an isomorphism in cohomology $f^*: H^*(N) \rightarrow H^*(M)$. Is it true then that $f$ induces an isomorphism in equivariant cohomology $f^*: H_G^*(N) \rightarrow H_G^*(M)$? If not, is it true if $f$ is a deformation retract?
I know that equivariant cohomology of a $G$-space $X$ can be computed as the cohomology of the Cartan complex $$ ( (S^*(\mathfrak{g}^\vee) \otimes_\mathbb R \Omega^*(X))^G, d_G)$$ where $d_G$ is the Cartan differential. I'm assuming that $f^*:H^*_G(N) \rightarrow H_G^*(M)$ can be seen as descending from the map: $$ 1 \otimes f^*: S^*(\mathfrak{g}^\vee) \otimes_\mathbb R \Omega^*(N) \rightarrow S^*(\mathfrak{g}^\vee) \otimes_\mathbb R \Omega^*(M) $$ where in $1 \otimes f^*$, $f^*$ just means ordinary pullback of forms. Is this assumption correct, and if so, how can it be used to show that $f^*: H_G^*(N) \rightarrow H_G^*(M)$ is an isomorphism, if this is true at all?
The particular case I'm interested in is that of $N$ being a submanifold of some manifold and $M$ being an equivariant tubular neighborhood $U$ of $N$. Then taking $\phi: U \rightarrow \nu_N$ to be an equivariant diffeomorphism mapping $N$ to the zero section of the normal bundle $\nu_N$ and taking $\pi: \nu_N \rightarrow N$ to be the projection, we have that $$\phi^* \pi^*: H^*(N) \rightarrow H^*(U)$$ is the identity map in cohomology by the invariance of de Rham cohomology because it comes from the deformation retract $\pi \circ \phi: U \rightarrow \nu_N \rightarrow N$. I am thus trying to conclude that: $$\phi^* \pi^*: S^*(\mathfrak{g}^\vee) \otimes_\mathbb R \Omega^*(N) \rightarrow S^*(\mathfrak{g}^\vee) \otimes_\mathbb R\Omega^*(U) $$ induces the identity in equivariant cohomology.