Isomorphism between algebras over C I would like to know how to find isomorphisms (or simply morphisms) $ \varphi : \mathcal{M}_{3} ( \mathbb{C} ) \to \mathcal{M}_3 ( \mathbb{C} ) $ of $\mathbb C$-algebras  which respect the following condition :
$ \forall a,b,c \in \mathbb{C} $, 
$$ \varphi \Big( \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix} \Big) = \begin{pmatrix} 0 & b & 0 \\ 0 & 0 & a \\ c & 0 & 0 \end{pmatrix} .$$
In other words, how to determine   $ \varphi \Big( \begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{pmatrix} \Big)$ ?
Thanks a lot.  :-)
 A: Let us prove that if $T:M_{n}(\mathbb{C})\rightarrow M_{n}(\mathbb{C})$ is an isomorphism of algebras then exists $A\in M_{n}(\mathbb{C})$ such that $T(X)=AXA^{-1}$.
Let $M\in M_{n}(\mathbb{C})$. Consider the linear transformation $S_{M}(X)=MX$ and notice that the rank of $S$ is $n\times \text{rank}\ M$. 
Let $T:M_{n}(\mathbb{C})\rightarrow M_{n}(\mathbb{C})$ be an isomorphism of algebras. 
Consider the linear transformation $T(S(T^{-1}(X)))=T(MT^{-1}(X))=T(M)X$. 
Notice that the rank of $TST^{-1}$ is the rank of $S$.
Notice also that $T(S(T^{-1}(X)))=S_{T(M)}$, therefore the rank of $TST^{-1}$ is $n\times \text{rank}\ T(M)$. 
So rank$(M)=$rank$(T(M))$. Thus, $T$ is a linear tranformation that preserves rank.
Marcus and Moyls proved in http://msp.org/pjm/1959/9-4/pjm-v9-n4-p17-s.pdf (See corollary of theorem 1) that a rank preserver transformation must be $T(X)=AXB$ or $T(X)=AX^tB$, where $A$ and $B$  are invertible matrices.
Since $T(Id)=Id$, because $T$ is an algebra isomorphism, we get $B=A^{-1}$. 
Now, if $T(X)=AX^tA^{-1}$ then $T(CB)=AB^tC^tA^{-1}$. 
But $T(CB)=T(C)T(B)=AC^tA^{-1}AB^tA^{-1}=AC^tB^tA^{-1}$.
So $CB=BC$ for every $C,B$. It is a contradiction. Thus, $T(X)=AXA^{-1}$ is the only option.
Remark: If you want $T(Y)=Y($for some $Y)$, you must have $AYA^{-1}=Y$ therefore $AY=YA$. You just need to choose a matrix $A$ that commute with all $Y$ that you want. 
