The answer to the following question might be obvious but I haven’t found a full proof yet (neither by myself nor in the literature). So my apologies if it is trivial. 


Let $X$ be a (for simplicity quasi-projective and non-singular) complex variety $X$ on which a finite group $G$ acts. 
Deligne has shown in Hodge III that the equivariant cohomology group $H\_G^k(X,\mathbb{Q})$ for any $k$ carries a natural mixed Hodge structure (MHS). On the other hand, the Leray-Serre spectral sequence for the Serre fibration 
\begin{equation}
X \to X\times\_G EG \to BG
\end{equation}
degenerates (complex topology) over $\mathbb{Q}$ and hence yields an isomorphism 


\begin{equation}
H\_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G
\end{equation}


(cf. this MO question: [Equivariant cohomology of finite group actions and invariant cohomology classes](https://mathoverflow.net/questions/120447/equivariant-cohomology-of-finite-group-actions-and-invariant-cohomology-classes?rq=1)). 
Clearly, $H^k(X,\mathbb{Q})^G$ inherits a MHS from the MHS on $H^k(X,\mathbb{Q})$. 
Now my question is: 


Are these two MHS on $H\_G^k(X,\mathbb{Q})$ naturally isomorphic?
More precisely, is there a simplicial version of the above Serre fibration yielding an isomorphism $H\_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G$ of MHS?


A natural candidate for such a simplicial version is 


\begin{equation}
[X/G]\_\bullet \to B\_\bullet G,
\end{equation}


where I (essentially) use Deligne's notation from Hodge III. Since I'm still learning simplicial methods, I was not sure if this is maybe too naive. 


Any thoughts/references/comments are very welcome!



Two mixed Hodge structures on equivariant cohomology for actions by finite groups
As I said in my comment, the mixed Hodge structures are the same. Here is the outline. From [Hodge III, 6.1.2.1], 
$$[X/G]\_n = (G^{n+1}\times X)/G$$
One has a descent spectral sequence
$$E\_1= H^q([X/G]\_p,\mathbb{Q})\cong (G^{p+1}\times H^q(X,\mathbb{Q}))/G$$
abutting to $H^{p+q}([X/G]\_\bullet, \mathbb{Q})=H^{p+q}\_G(X,\mathbb{Q})$ [Hodge III, (5.2.1.1)], and this is compatible with MHS [Hodge III, (8.1.15)]. Now use the fact that the complex $E\_1$ is the bar complex, which computes group cohomology; in this case it is trivial except in degree zero. So in conclusion
$$H\_G^\*(X,\mathbb{Q})\cong H^\*(X,\mathbb{Q})^G$$
as MHS.


