Rewriting automorphism of matrix algebra in terms of automorphisms of the underlying ring? I've used the following idea as a black box for some time now, but it occurred to me I don't fully understand why it's true.
Suppose $A=M_n(R)$ is the algebra of square matrices over some division ring $R$. Then for any $\phi\in\operatorname{Aut}(A)$, we can actually write $\phi$ as the composition of an automorphism induced by an automorphism $\psi$ of $R$ and the conjugation by some unit of $A$. 
More explicitly, for $\psi\in\operatorname{Aut}(R)$, this induces an automorphism $\tilde{\psi}$ of $A$ by applying $\psi$ to each of the entries in the matrix, for example,
$$
M=\begin{pmatrix}
a_{11} & a_{12}\\ a_{21} & a_{22}
\end{pmatrix}
\mapsto
\tilde{\psi}(M)\begin{pmatrix}
\psi(a_{11}) & \psi(a_{12})\\ \psi(a_{21}) & \psi(a_{22})
\end{pmatrix}
$$
and then we can conjugate by an invertible matrix in $A$, say $N$, to get $N\tilde{\psi}(M)N^{-1}$. I don't think the order of applying $\tilde{\psi}$ or conjugating matters, since if I conjugate first, then I could apply a different $\tilde{\psi}$. So the composition would be something like $\phi=\varphi_N\circ\tilde{\psi}$ where $\varphi_N$ is the conjugation by $N$ map. 
My question is, why can any automorphism $\phi$ of $A$ actually be decomposed in this way?
 A: A special case: Let $\phi:M_n(R)\to M_n(R)$ be an automorphism. It restricts to an automorphism of the center of $M_n(R)$, which is the same as the center $K=Z(R)$ of $R$, which is a field. If we suppose that this restriction $\phi|_K$ is the identity and that $R$ is finite, then the Nother-Skolem theorem tells us that $\phi$ is inner, that is, by conjugation by an invertible element of $M_n(R)$.
The general case is stated in Algebra IX: Finite Groups of Lie Type, Finite-dimensional Division Algebras, by A. I. Kostrikin and I. R. Shafarevich, in chapter II, section 3. They see $M_n(R)$ as an algebra over a field and look for automorphisms which are algebra automorphisms: but you can always take the ground field to be the prime field of the center of $R$.
A: I think what you're describing is basically the contents of the Skolem-Noether theorem which states that a central simple algebra $A$ with center $Z$ has the property that every $Z$-algebra automorphism is inner.
So in the case of a matrix ring over a field $\mathbb{F}$, all $F$-linear automorphisms are inner. 
However if your automorphism may not be $R$-linear and it may not be central if you use a division ring $R$, that is, the map doesn't fix $R$. Here you could say: "Well, $\phi(R)$ is isomorphic to $R$, so why don't I just identify $R$ with $\phi(R)$ and pretend the original map is $R$ linear?"
This sounds a little bit like what you described, although it's beyond what I've seen stated with the Skolem-Noether theorem.
