Turning my short sloppy proof that $G \blacktriangleright H \blacktriangleright K$ then $G \blacktriangleright K$ into a concise and neat proof?

Question

Continuing to work on writing clean proofs while studying for a qualifying exam (1st try as I enter grad school so there are no stakes involved). I can "see" the rationale for the titular proof in my head and can explain it (sort of) in words but I'm at a loss for the cleanest and most concise way to write it out. Here's my attempt...

$\mathbf{Theorem}$: group $G$, and subgroups $H$ and $K$ which form a "chain" of characteristic subgroups $G \blacktriangleright H > \blacktriangleright K$ show that $G \blacktriangleright K$.

$\mathbf{Proof}$: The groups of automorphisms on $G, H$, and $K$ are denoted as $\operatorname{Aut}(G), \operatorname{Aut}(H)$, and $\operatorname{Aut}(K)$ respectively.

Every element has $\operatorname{Aut}(H)$ is "embedded" in some element of $\operatorname{Aut}(G)$ because $H$ is characteristic in $G$.

Similarly every element of $\operatorname{Aut}(K)$ is "embedded" in some element of $\operatorname{Aut}(H)$.

Since $K \blacktriangleleft H$ we have every element of $\operatorname{Aut}(H)$ has some element of $\operatorname{Aut}(K)$ embedded in it.

Smiilarly, since $H \blacktriangleleft G$ we have every element of $\operatorname{Aut}(G)$ has some element of $\operatorname{Aut}(H)$ embedded in it.

Clearly we have that every automorphism of $G$ has an automorphism of $K$ embeded in it. Hence $K \blacktriangleleft G$, $\square$.

$\mathbf{Self Critique}$: I believe that speaking of "embedding" without being very clear about what that means is a problem but am unsure how to flesh it out for precision's sake (and possibly that might make this proof inaccurate?)

It was suggested that this is a duplicate of Being characteristic is transitive which it is not. This question is tagged as "proof-writing" as in I needed feedback in writing a better version of my own proof and not simply copying another person's proof.

Answer 1

Any automorphism of $G$ fixes $H$. Any automorphism of $G$, restricted to $H$, is an automorphism of $H$. Any automorphism of $H$ fixes $K$, so any automorphism of $G$ fixes $K$. Therefore $K$ is characteristic in $G$.