It is well-known that if $O$ is an orthogonal map, then $\Delta u(Ox) = \Delta u$ where $\Delta$ is the Laplacian.
Now, let $A$ be a constant invertible matrix, then we define the weighted Laplacian
$$\Delta_A = \langle \nabla, A \nabla \rangle.$$
My question is: Does there exist a function $\varphi_O$ such that $Gf(x):=e^{\varphi_O(x)}f(Ox)$ satisfies for all $f$ but fixed orthogonal $O$
$$\Delta_A Gf(x) = G(\Delta_A f)(x)?$$
$\newcommand{\De}{\Delta}$The answer is: not in general.
Indeed, suppose the contrary. Let \begin{equation} A:=\begin{pmatrix} 0&1/2\\ 1/2&0 \end{pmatrix}, \end{equation} so that \begin{equation} (\De_A f)(s,t)=\frac{\partial^2 f(s,t)}{\partial s\,\partial t}=f^{(1,1)}(s,t). \end{equation} Let \begin{equation} O:=\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} \end{equation} and \begin{equation} g:=\varphi_O, \end{equation} so that \begin{equation} (Gf)(s,t):=e^{g(s,t)}f(-t,s), \end{equation} and the difference between the right-hand side and left-hand side of the equality in question divided by $e^{g(s,t)}$ is \begin{equation} \begin{aligned} d_f(s,t)&:=\frac{(G(\De_A f))(s,t)-(\De_A(Gf))(s,t)}{e^{g(s,t)}} \\ &=\frac{e^{g(s,t)}f^{(1,1)}(-t,s) -\dfrac{\partial^2 f(s,t)}{\partial s\,\partial t} \big(e^{g(s,t)}f(-t,s)\big)}{e^{g(s,t)}} \\ &=-f^{(0,1)}(-t,s) g^{(0,1)}(s,t)+f^{(1,0)}(-t,s) g^{(1,0)}(s,t) +2 f^{(1,1)}(-t,s) \\ &-f(-t,s) \left(g^{(0,1)}(s,t) g^{(1,0)}(s,t)+g^{(1,1)}(s,t)\right)=0 \end{aligned} \end{equation} for all $f\in C^1$ and all real $s,t$.
Letting now $f_1(s,t):=s$, $f_2(s,t):=t$, and $f_4(s,t):=t^2$, we get \begin{equation} 0=s d_{f_2}(s,t)-d_{f_4}(s,t)=s g^{(0,1)}(s,t), \end{equation} so that $g^{(0,1)}=0$ and hence $g^{(1,1)}=0$, which implies $0=d_{f_1}=g^{(1,0)}$. So, $g$ is a constant, and hence $0=d_f(s,t)=2f^{(1,1)}(-t,s)$ for all $f\in C^1$ and all real $s,t$, which is absurd. $\quad\Box$
