If $\pi $ is a permutation representation, then $\pi (\text{Hol} (H))=N_{S_n}(\pi(H))$

Question

Let $|H|=n$, $K=\text{Aut} (H)$ and $G=\text{Hol}(H)$. Let $G$ act on the cosets of $K$ and use this action to define the faithful representation $\pi : G\to S_n$. Prove $\pi (G)=N_{S_n}(\pi(H))$. It is easy to prove $\pi (G) \le N_{S_n}(\pi(H))$, the text gives the hint "show $|G|=|N_{S_n}(\pi(H))|$ using the fact that for a semidirect product $A\rtimes _{\phi} B$, $C_B(A)=\text{ker} \phi$ and $C_A(B)=N_A(B)$".
I found a solution that uses the orbit stabilizer theorem to show that $|N_{S_n}(\pi(H))|=n|\text{stab} K|$ which follows because $N_{S_n}(\pi(H))$ is transitive on the set of cosets of $K$ and then shows that $|\text{stab} K| \le |\pi (K)|$ by finding an injection from the former to the latter.
However, nowhere in the proof is the second fact given in the hint used, which makes me believe that there must be a quicker way to solve the problem. Yet, I have been unable to use this second fact and I wanted to know if a proof using it would indeed be nicer or if it would turn out to be more convoluted.