Let $M = GL(n,\mathbb{C})$ be the set of nondegenerate bilinear forms on $\mathbb{C}^n$ (not necessarily symmetric). The general linear group $G = GL(n,\mathbb{C})$ acts on $M$ in the usual way $$G\times M \rightarrow M,~~~~~~(A,X) \mapsto AXA^T.$$ What is the equivariant cohomology $H^*_G(M)$?

2$\begingroup$ I mildly object to denoting the set of nondegenerate bilinear forms $GL(m,\mathbb C)$. :/ $\endgroup$ – Mariano SuárezÁlvarez Jul 5 '13 at 22:03
I'll assume that you mean the Borel cohomology $H^*(EG\times_GM)$. If so, the answer is
$$ \mathbb{Z}[\![c_1,\dotsc,c_n]\!]\otimes\Lambda^*(a_1,\dotsc,a_n), $$
where $c_k=2k$ and $a_k=2k1$. To see this, first note that you can replace everything by maximal compact subgroups without changing the homotopy type. This means that the relevant space is $EU(n)\times_{U(n)}U(n)^{\text{ad}}$, where $U(n)^{\text{ad}}$ refers to the space $U(n)$ with $U(n)$ acting on it by conjugation. This can be described in a different way as follows. For any complex vector bundle $V$ (with Hermitian inner product) over a space $X$, we can consider the fibre bundle $U(V)$ whose points are pairs $(x,g)$, where $x\in X$ and $g$ is a unitary automorphism of $V_x$. If we take $V$ to be the tautological bundle over $BU(n)$ then $U(V)$ is easily identified with $EU(n)\times_{U(n)}U(n)^{\text{ad}}$. For any $V$ and $X$ it is known that
$$ H^*(U(V)) = H^*(X) \otimes\Lambda^*(a_1,\dotsc,a_n), $$
and my claim above is a special case of this. The key ingredient in the proof is Miller's stable splitiing theorem for $U(n)$, in the equivariant form proved by Nitu Kitchloo; some additional details are in my paper "Common subbundles and intersections of divisors".

$\begingroup$ Hi Neil, thanks for your answer. Unfortunately it is not quite what I am looking for. My action is $(A,X) \mapsto AXA^T$ where $A^T$ is the regular transpose, not the conjugate transpose. Restricting to the unitary group, the action is $(A,X) \mapsto A X A^T = A X \bar{A}^{1}$, where the bar means entrywise complex conjugation. To illustrate in the case $n=1$, the adjoint action of $U(1)$ on $U(1)$ is trivial, but the action I'm thinking about is by $(a,x) \mapsto axa = a^2x$ which is transitive with stabilizer $\pm 1$. The homotopy quotient is then $BZ_2 = \mathbb{R} P^{\infty}$. $\endgroup$ – Tom Baird Jul 7 '13 at 19:22
Update: I managed to partially answer this question after posing it. For coefficient fields relatively prime to $n$!, there is an isomorphism $ H_G^*(M) \cong H^*(BLO(n))$ where $BLO(n)$ is the classifying space of the continuous loop group $LO(n)$, while in characteristic 2 there is an isomorphism $ H_G^*(M) \cong H^*(BLU(n))$.
Details are found in Corollary 6.4 of http://front.math.ucdavis.edu/1312.7450.