Wikipedia Proof of Skolem-Noether Theorem Could someone help me understand the wikipedia proof of the Skolem-Noether theorem? In particular, I'm interested in the case where $A=B= M_n(\mathbb{C})$.

So the theorem claims that given any two homomorphisms $f, g: A \to A$, there is a unit $k \in A$ so that for each $a$, $f(a)=kg(a)k^{-1}$.

It seems this case is dealt with in the first two sentences of the wikipedia proof. They observe that $f, g$ give an action of $A$ on $\mathbb{C}^n$ via $a \cdot v=f(a)v$. This makes $\mathbb{C}^n$ into $A$ modules in two different ways which we can denote $V_f, V_g$. Now, there is a unique simple $A$-module $S$. Thus, $V_f \cong S^r, V_g \cong S^s$ for some $r ,s$.
If $V_f$ and $V_g$ are isomorphic as $A$-modules, then such a map is also an isomorphism of $\mathbb{C}$-vector spaces and so there is some invertible matrix $b: V_f  \to V_g$ and I am done. But this is seemingly only true if $s=r$. What am I missing?
Edit: This is the case since $V_f, V_g$ have the same dimension as vector spaces and so the dimensions of $S^r, S^s$ are the same.
 A: 
Let $A=M_n(\mathbb{C})$ and $f, g: A \to A$ be $\mathbb{C}$-algebra homomorphisms. I will prove that there is some invertible matrix $k \in A$ so that for all $a \in A$, $kf(a)k^{-1}=g(a)$.

If we consider that vector space $\mathbb{C}^n$, then $f, g$ each give an $A$-action on $\mathbb{C}^n$ by $a \cdot v=f(a)v, a \cdot v = g(a)v$ respectively for $a \in A, v \in \mathbb{C}^n$. Thus, we denote the resulting $A$-modules as $V_f, V_g$. Since $f, g$ are $\mathbb{C}$-algebra homomorphisms, they preserve the action of $\mathbb{C}$. Since $\mathbb{C} \subset A$ and $V_f, V_g$ are finite dimensional vector spaces, they must be finitely generated $A$-modules.
We remark that if $0 \to N \to M \to M/N \to 0$ is exact and $M$ is semisimple, then so are $N$ and $M/N$. Now, $V_f, V_g$ are quotients of free $A$-modules and a finitely generated free $A$-module is just some power of $A$. Thus, if we can show that $A$ is semisimple over itself, then so are $V_f, V_g$.
But $A$ is a direct product of the submodules $C_i$ where only the $i$th column is non-zero and these are simple since any non-zero vector of $C_i$ generates all of $C_i$.
Moreover, any simple $A$-module is a quotient of $A$ by some maximal left ideal and such quotients are all isomorphic to the $C_i$. Thus, up to isomorphism, there is only one simple $A$-module and we denote it $S$.
So since $V_f, V_g$ are semisimple, they are isomorphic to $S^r, S^s$ respectively. Now, $S$ has dimension $n$ as a $\mathbb{C}$-vector space, and so do $V_f, V_g$. So, $r=s=1$ and so we conclude that $V_f, V_g$ are isomorphic as $A$-modules. Such an isomorphism is also an isomorphism of $\mathbb{C}$-vector spaces and so is determined by some invertible matrix $k$. Thus, for any vector $v \in \mathbb{C}^n$ and $a \in A$, we have $k(f(a)v)=k (a \cdot v)=a \cdot k(v)=g(a)k(v)$. This holds for all $v$ and so $k f(a)=g(a)k$ as desired.
