$\mathbb{C}$-algebra automorphism of $M_n(\mathbb{C})$ has form $X \mapsto AXA^{-1}$. As the title suggests, what is the easiest way to see that any $\mathbb{C}$-algebra automorphism of $M_n(\mathbb{C})$ has the form $X \mapsto AXA^{-1}$ for some fixed $A \in GL_n(\mathbb{C})$?
 A: It is a consequence of Skolem-Noether theorem.
A: I don't really know what "easiest" means: this heavily depends on what you are taking as granted.
Anyway, here is a direct elementary proof. In what follows, $\Phi$ is an automorphism of $M_N(\mathbb C)$.
Denote by $(E_{i,j})_{1\leq i,j\leq N}$ the "canonical basis" of $M_N(\mathbb C)$. If $(e_1,\dots ,e_N)$ is the canonical basis of $\mathbb C^N$, then
$$E_{i,j}e_k=\delta_{j,k}e_i\qquad\hbox{for $i,j,k=1\dots N$} ,$$
where $\delta_{j,k}$ is the usual "Kronecker symbol". Note that
$$e_k=E_{k,1}e_1\qquad\hbox{for $k=1,\dots ,N$}. $$
Set also 
$$F_{i,j}:=\Phi(E_{i,j})\, .$$
By definition, the $E_{i,j}$'s satisfy the identities 
$$E_{i,j}E_{k,l}=\delta_{j,k}E_{i,l}\, .$$
Since $\Phi$ is an algebra homomorphism, the $F_{i,j}$'s satisfy the same identities, and they are non-zero because $\Phi\neq 0$. In particular, $F_{k,k}$ is a non-zero projection for every $k$.
Choose any non-zero vector $f_1$ in the range of $F_{1,1}$. Then $F_{1,1}f_1=f_1$ because $F_{1,1}$ is a projection. Now, define 
$$ f_k:=F_{k,1}f_1\qquad\hbox{for $k=1,\dots ,N$.}$$
Note that $f_1,\dots ,f_N$ are linearly independent. Indeed, assume that
$$\sum_{j=1}^N\lambda_j f_j=0 $$
for some scalars $\lambda_1,\dots ,\lambda_N$. Applying $F_{1,k}$
 and since $f_j=F_{j,1}f_1$ and $F_{1,k}F_{j,1}=\delta_{j,k}F_{1,1}$ for all $j$, we get
$$0=\lambda_kF_{1,1}f_1=\lambda_kf_1\, , $$
so that $\lambda_k=0$ for $k=1,\dots ,N$.
So we know that $(f_1,\dots ,f_N)$ is a basis of $\mathbb C^N$. Let $P$ be the invertible matrix such that 
$$f_k=Pe_k\qquad\hbox{for $k=1,\dots ,N$.} $$
Denote by $\Phi_P$ the automorphism of $M_N(\mathbb C)$ defined by $\Phi_P(X)=PXP^{-1}$. Let us show that $\Phi=\Phi_P$.
To do this, it is enough to check that $\Phi(E_{i,j})=\Phi_P(E_{i,j})$ for all $i,j\in\{ 1,\dots ,N\}$; and this can be done by testing on the vectors $f_k$. On the one hand, 
$$\Phi(E_{i,j})f_k=F_{i,j}f_k=F_{i,j}F_{k,1}f_1=\delta_{j,k}F_i,1f_1=\delta_{j,k}f_i\, ; $$
and on the other hand
$$\Phi_P(E_{i,j})f_k=PE_{i,j}P^{-1}f_k=PE_{i,j}e_k=\delta_{j,k}Pe_i=\delta_{j,k}f_i\, . $$
So we have indeed $\Phi=\Phi_P$.
A: Here is a simple direct proof.  Suppose $f:M_n(\mathbb{C})\to M_n(\mathbb{C})$ is an automorphism.  For $1\leq i,j\leq n$, let $e_{ij}$ denote the matrix with $ij$ entry $1$ and all other entries $0$, and write $f_{ij}=f(e_{ij})$.  Note that $e_{ij}e_{k\ell}=\delta_{jk}e_{i\ell}$, and hence by applying $f$ to both sides we have $f_{ij}f_{k\ell}=\delta_{jk}f_{i\ell}$.
For each $i$, let $V_i\subseteq \mathbb{C}^n$ be the image of $f_{ii}$.  Consider the map $F:\mathbb{C}^n\to \bigoplus_{i=1}^n V_i$ given by $F(v)=(f_{11}v,\dots,f_{22}v)$ and the map $G:\bigoplus_{i=1}^n V_i\to \mathbb{C}^n$ given by $G(v_1,\dots,v_n)=\sum_{i=1}^n v_i$.  Since $\sum f_{ii}=I$, $GF=I$.  On the other hand, the identities $f_{ii}^2=f_{ii}$ and $f_{ii}f_{jj}=f_{jj}f_{ii}=0$ for $i\neq j$ imply
$$FG(f_{11}w_1,\dots,f_{nn}w_n)=(f_{11}\sum f_{ii}w_i,\dots,f_{nn}\sum f_{ii}w_i)=(f_{11}w_1,\dots,f_{nn}w_n),$$
so $FG=I$ as well.  Thus $G$ is an isomorphism, which says exactly that $\mathbb{C}^n$ is the internal direct sum of the subspaces $V_i$.  Since each $V_i$ is nonzero (since $f_{ii}$ must be nonzero) and there are $n$ of them, this implies each $V_i$ must be 1-dimensional.
Now fix a nonzero vector $v_1\in V_1$, and for $i>1$, define $v_i=f_{i1}v_1$.  Since $f_{ij}f_{k\ell}=\delta_{jk}f_{i\ell}$, we have $$f_{ij}v_k=\delta_{jk}v_i$$ for all $i,j,k$ (for $i=1$, you must use the fact that $f_{11}v_1=v_1$ since $v_1\in V_1$).  In particular, setting $i=j=k$, this implies $v_k\in V_k$ for each $k$, and setting $j=k$ and $i=1$, each $v_k$ is nonzero.  Since $\mathbb{C}^n=\bigoplus V_i$ and each $V_i$ is 1-dimensional, this means $\{v_1,\dots,v_n\}$ is a basis for $\mathbb{C}^n$.  Let $A\in M_n(\mathbb{C})$ be the matrix whose columns are the $v_i$.  Then it is easy to compute that $$Ae_{ij}A^{-1}v_k=\delta_{jk}v_i.$$  Since the $v_k$ form a basis, this implies $Ae_{ij}A^{-1}=f_{ij}=f(e_{ij})$.  Since the $e_{ij}$ span all of $M_n(\mathbb{C})$, it follows that $f(X)=AXA^{-1}$ for all $X\in M_n(\mathbb{C})$.
