I think the answer to your question is negative. Consider for instance $$M = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), \quad N = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$ Assume that there exist $$A = \left(\begin{array}{cc} a_{11} & a_{12} \\ 0 & a_{22} \end{array}\right), \quad B = \left(\begin{array}{cc} b_{11} & b_{12} \\ 0 & b_{22} \end{array}\right)$$ with $det(A) = det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \left(\begin{array}{cc} a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{array}\right) = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $det(A) = det(B) = 1$.

More generally, your action stabilizes the locus of matrices of the form $\left(\begin{array}{cc} m_{11} & m_{12}\\ m_{21} & 0 \end{array}\right)$.

I think the answer to your question is negative. Consider for instance $$M = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ Assume that there exist $$A = \begin{pmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{pmatrix}, \quad B = \begin{pmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{pmatrix}$$ with $\det(A) = \det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \begin{pmatrix} a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $\det(A) = \det(B) = 1$.

More generally, your action stabilizes the locus of matrices of the form $\begin{pmatrix} m_{11} & m_{12}\\ m_{21} & 0 \end{pmatrix}$.

I think the answer to your question is negative. Consider for instance $$M = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), \quad N = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$ Assume that there exist $$A = \left(\begin{array}{cc} a_{11} & a_{12} \\ 0 & a_{22} \end{array}\right), \quad B = \left(\begin{array}{cc} b_{11} & b_{12} \\ 0 & b_{22} \end{array}\right)$$ with $det(A) = det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \left(\begin{array}{cc} a_{11}b_{11}+a_{12}b_{22} & a_{11}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{array}\right) = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $det(A) = det(B) = 1$.

More generally, your action stabilizes the locus of matrices of the form $\left(\begin{array}{cc} m_{11} & m_{12}\\ m_{21} & 0 \end{array}\right)$.

I think the answer to your question is negative. Consider for instance $$M = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), \quad N = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$ Assume that there exist $$A = \left(\begin{array}{cc} a_{11} & a_{12} \\ 0 & a_{22} \end{array}\right), \quad B = \left(\begin{array}{cc} b_{11} & b_{12} \\ 0 & b_{22} \end{array}\right)$$ with $det(A) = det(B) = 1$ and such that $$A\cdot M\cdot B^{T} = N$$ Then $$A\cdot B^{T} = \left(\begin{array}{cc} a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ a_{22}b_{12} & a_{22}b_{22} \end{array}\right) = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $det(A) = det(B) = 1$.

More generally, your action stabilizes the locus of matrices of the form $\left(\begin{array}{cc} m_{11} & m_{12}\\ m_{21} & 0 \end{array}\right)$.