For the group $GL(3,\mathbb R)$ acting on $M_{3}(\mathbb R)$ by $A\cdot M=AM$. Let $$M_1 = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix}$$ $$M_2 = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 2 & 2 \end{pmatrix}$$
I have to find out if M1 or M2 lie in the same orbit, I understand that I may need to determine:$$\text{Stab}(M_1)=A\text{Stab}(M_2)A^{-1}$$ However, I am stuck on how I can find/identify the stabilizer and the matrix A, can anyone offer some guidance?
Suppose
$$\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}1&2&0\\0&0&0\\0&1&2\end{pmatrix}=\begin{pmatrix}1&2&0\\0&0&0\\0&2&2\end{pmatrix}\iff$$
$$\begin{pmatrix}a&2a+c&2c\\d&2d+f&2f\\g&2g+i&2i\end{pmatrix}=\begin{pmatrix}1&2&0\\0&0&0\\0&2&2\end{pmatrix}$$
We get:
$$a=1,\,c=0\;,\;\;d=f=0\;,\;\;g=0\;,\;\;\text{and now the problem:}$$
From entry $\;32\;$ we get that $\;2g+i=i=2\implies i=2\;$ , but then entry $\;33\;$ gives us $\;2i=2\implies i=1\;$ , and we get a contradiction.