Let $A$, $B$ be two commutative and associative $\mathbb k$-algebras and let $A_\hbar:=A[[\hbar]]$, $B_\hbar:=B[[\hbar]]$ be the corresponding ring of formal series. Of sense [Deformation theory and quantization I; F. Bayen, M. Flato, C. Frønsdal, A. Lichnerowicz, D.Sternheimer; 1978] or [Deformation quantization of Poisson manifolds, I; M. Kontsevich; 1997; http://arxiv.org/abs/q-alg/9709040] we consider two star products $*_1$ and $*_2$ on $A$ and $B$. Then $(A_\hbar, *_1)$ and $(B_\hbar, *_2)$ are two associative $\mathbb k [[\hbar]]$-algebras. One can say that $*_1$ and $*_2$ are "weak equivalents" if there is an algebra $\mathbb k [[\hbar]]$-isomorphism between $(A_\hbar, *_1)$ and $(B_\hbar, *_2)$. We also can say that $*_1$ and $*_2$ are "equivalents" if there is a sequence $\{T_r:A\to B\}_{r\in \mathbb N}$ of bijective $\mathbb k$-maps such that

i) $T_0:A\to B$ is an algebra isomorphism.

ii) For all $a,b\in A_\hbar$, $T(a *_1 b)=T(a)*_2T(b)$, where $T:=T_0+T_1\hbar+T_2\hbar^2+\cdots$

It is clear (for me I guess) that "equivalent" implies "weak equivalent" and if we have that $A=B$, then the notion of "equivalent" coincide with the usual one given by [BFFLS].

I have two questions:

1. "Weak equivalence" implies "equivalence" ?

2. If a Lie group $G$ acts on $A$ and $B$, so this action extend $\mathbb k[[\hbar]]$-linearly on $A_\hbar$ and $B_\hbar$. Suppose that $*_1$ and $*_2$ are equivalents, $T:(A_\hbar,*_1)\to (B_\hbar,*_2)$, and suppose also that $T$ is a $G$-morphism. What we can say about the $T_r$ of $T$ ? Are they $G$-morphisms ?

Any comment(s) would be highly appreciated. I'll be so glad if you know and point me some references about this subject. Thank you in advance.