Finding invertible 3x3 matrix A such that Stab(M)A=AStab(N) for given matrices M,N I'm reading that it is possible to find an invertible $3 \times 3$ matrix $A$ such that for the matrices
$M = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ and 
$N = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix}$
we have $\operatorname{Stab}(M).A = A.\operatorname{Stab}(N)$.
Here $\operatorname{Stab}(M) = \left\{ B \in \operatorname{GL}(3,\mathbb{R}) : BM = M \right\}$
I believe it's possible but I can't find it at all!
I've tried right multiplying the equality by $N$ then I get $B.A.N = A.C.N =A.N$ (for any $B \in \operatorname{Stab}(M)$ and the corresponding $C \in \operatorname{Stab}(N)$) so that $\operatorname{Stab}(M)$ is actually a subset of $\operatorname{Stab}(A.N)$. Also right multiplying by $A^{-1}$ gives $\operatorname{Stab}(M) = A.\operatorname{Stab}(N).A^{-1}$ then right multiplying by $M$ and left multiplying by $A^{-1}$ again gives $A^{-1}.M = \operatorname{Stab}(N).A^{-1}$.
Do you have a solution to this? Am I actually on the right path? Do you have other ideas?
Thank you so much!

Your ideas would be good if this was a general property of $M$ and $N$, but I don't think it is. You need to use the specific $M$ and $N$.
The names of these two ideas are “orbit-stabilizer” and “Gauss–Jordan elimination.”
GJ:
Notice that $\begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. N
=\begin{bmatrix} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix}$,
and so
$\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot  \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot  N
= \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}
$
and so
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}
\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot  \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot  N
= \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} = M
$
Setting $A = 
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}
\begin{bmatrix} \tfrac 12 & 0 & 0 \\ 0 & \tfrac 12 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\cdot  \begin{bmatrix} 1& 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
= \begin{bmatrix} \tfrac12&0&\tfrac12\\ 0&\tfrac12&0\\-\tfrac12&-\tfrac12&\tfrac12\\ \end{bmatrix}
$
We have $M=A.N$.
OS:
$C.A \in \operatorname{Stab}(M).A$ iff $C \in \operatorname{Stab}(M)$ iff $C.M=M$ iff $C.(A.N) = (A.N)$ iff $\left(A^{-1}. C .A\right).N = N$ iff $\left( A^{-1}.C.A \right) \in \operatorname{Stab}(N)$ iff $A.\left(A^{-1}.C.A\right) = C.A \in A.\operatorname{Stab}(N)$.
In other words, $X=C.A$ is in $\operatorname{Stab}(M).A$ if and only if $X$ is in $A.\operatorname{Stab}(N)$. So  $\operatorname{Stab}(M).A= A.\operatorname{Stab}(N)$.
