Isomorphism of subalgebras

Question

If $A$ and $B$ are isomorphic subalgebras of $C$, is there necessarily an automorphism $\phi : C \rightarrow C$ mapping $A$ to $B$?

Can restrict this to a matrix algebra if it makes things more 'concrete'. Thanks!

Answer 1

Negative. Let $C=\Bbb C^{4\times 4}$ and note that each automorphism of $C$ is an inner automorphism. Now, take two matrices $a,b\in C$ which have the same minimal polynomial but who are not similar. For example, take \begin{align*} a &= \begin{pmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{pmatrix}, & b &= \begin{pmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}, \end{align*} who share the minimal polynomial $\mu=t^2$. Let $A:=\Bbb C[a]$ and $B:=\Bbb C[b]$. It is (in some cases) by definition that $(\mu)$ is precisely the kernel of the natural map $\Bbb C[t]\to A$ mapping $t\mapsto a$, so both $A$ and $B$ are isomorphic to $\Bbb C[t]/(t^2)$. More precisely, $A=\Bbb C e\oplus \Bbb C a$ and $B=\Bbb C e\oplus\Bbb C b$, where $e$ denotes the identity matrix.

If there was an automorphism $\phi:C\to C$ with $\phi(A)=B$, then it would have to map $a$ to (some scalar multiple of) $b$, which is impossible because there is no inner automorphism of $C$ which can achieve this: It would mean that $a$ and $b$ are similar.