Here $G$ is a group with subgroup $H$, and we let $G$ act on $G/H$ by left multiplication. Correspondingly, $G/H$ is a left $G$-set and the set $\operatorname{Aut}_G(G/H)$ denotes the set of all $G$-automorphisms in $G/H$.
There are already two posts discussing on this isomorphism (See this and this). Here for every $x\in N_G(H)$, we have a $G$-automorphism on $G/H$: $$\alpha_x:G/H\to G/H:aH\mapsto(ax^{-1})H,$$ and the following map is a group homomorphism $$\Phi:N_G(H)\to\operatorname{Aut}_G(G/H):x\mapsto\alpha_x.$$ It is clear that $\ker(\Phi)=H$, so the desired isomorphism is proven as long as we can show that $\Phi$ is surjective.
However, the arguments in the second post is false: Let $f$ be a $G$-automorphism on $G/H$ with $kH:=f(H)$. There is no guarantee that $k\in N_G(H)$ or $k^{-1}\in N_G(H)$, so $\alpha_k$ or $\alpha_{k^{-1}}$ need not be defined then.
I came up with one proof and want to check it here: Let $k'H:=f^{-1}(H)$. Note that $$H=f(f^{-1}(H))=f(k'H)=k'f(H)=(k'k)H,$$ so we have $k'k\in H$. By similar arguments, we have $kk'\in H$ as well.
Meanwhile, for every $h\in H$, we have $$kH=f(H)=f(hH)=hf(H)=(hk)H,$$ so $k^{-1}hk\in H$ holds for sure. In other words, we can see that $k^{-1}Hk\subseteq H$ now. Again, by similar arguments, we have $k'^{-1}Hk'\subseteq H$ as well, so $$kHk^{-1}=k(k'k)^{-1}H(k'k)k^{-1}=k'^{-1}Hk'\subseteq H,$$ implying that $H\subseteq k^{-1}Hk$ as well. Consequently, we have $H=k^{-1}Hk$ whence $k\in N_G(H)$. Then as in that post, we may notice that $f=\alpha_{k^{-1}}$, so the proof is complete.
P.S. It seems that $k^{-1}hk\in H$ for all $h\in H$ does not imply that $k\in N_G(H)$. I wonder if there were any nice counterexamples for that.
$\newcommand{\orb}{\mathsf{Orb}}$Your proof looks good to me.
Just for fun, consider the category $\orb'_G$ whose objects are subgroups of $G$ and arrows $H_1\to H_2$ are coset classes $[g]$ in $G/H_2$ with $H_1\le gH_2g^{-1}$, obvious identities and composition $[g'][g]:=[gg']$.
It can be shown using the orbit-stabiliser theorem that the functor from this category to the category of transitive left $G$-actions, $\orb_G$, which takes $H$ to the standard $G$-action on $G/H$ and $[g]:H_1\to H_2$ to the morphism $G/H_1\ni[x]\mapsto[xg]\in G/H_2$, is an equivalence. So it preserves automorphism groups. It follows that $[g]\mapsto([x]\mapsto[xg])$, which is a restatement of your $\phi$, makes a group isomorphism $(N(H)/H)^{\mathsf{op}}\cong N_{G^\mathsf{op}}(H)/H\cong\mathsf{Aut}(G/H)$ and composing with the inversion gives $N(H)/H\cong\mathsf{Aut}(G/H)$... which is exactly the isomorphism you propose. Here the right hand side stands for automorphisms of left $G$-actions.