TITLE: Equivariant Formality
QUESTION [5 upvotes]: Let $G$ be a finite group and $\mathcal{A}$ be a $dg$-algebra. Assume $G$ acts on $\mathcal{A}$, i.e. there exists a homomorphism $G\to {\rm Aut}_{dg}(\mathcal{A})$.
Assume further there exists a $dg$-algebra $\mathcal{B}$ and an isomorphism $f:\mathcal{A}\to \mathcal{B}$, then you can find a $G$-action on $\mathcal{B}$ such that $f$ is equivariant.

What is if $f$ not an isomorphism but only a quasi-isomorphism. Is it possible to find a $G$-structure on $\mathcal{B}$ such that $\mathcal{A}$ and $\mathcal{B}$ are equivariantly quasi-isomorphic.

I guess no. But I have hope that following statement is true:

If a $dg$-algebra $\mathcal{A}$ with a $G$-action is formal, then it is also equivariantly formal?  

Are any results in this direction known?

REPLY [3 votes]: I believe your second statement is true when $\mathcal{A}$ is the minimal Sullivan model of a $G$-space $X$. Namely, if the $G$-space $X$ is rationally formal, then it is rationally $G$-formal, in the sense that it has a rational $G$-minimal model which can be constructed from $H^\ast(X;\mathbb{Q})$ with it's $G$-action.
This is discussed in Section 3.3 of the book Algebraic Models in Geometry by Félix, Oprea and Tanré, see in particular Remark 3.30(2). They refer to a paper of John Oprea: Lifting homotopy actions in rational homotopy theory, 
J. Pure Appl. Algebra 32 (1984), no. 2, 177–190.