Firstly, I am aware this question has been asked on this site before. My issue is that I don't feel confident on my approach to the proof since other solutions use the concepts of conjugation.
Here is my approach to the proof. Assume $\phi:G\rightarrow H$ is an isomorphism. Let $\theta:{\rm Aut}(G) \rightarrow {\rm Aut}(H)$ by $\theta(\alpha)=\phi(\alpha)$, where $\alpha\in{\rm Aut}(G)$ be a function. I claim that $\theta$ is an isomorphism between ${\rm Aut}(G)$ and ${\rm Aut}(H)$.
Am I right?
In situations like this, it's often useful to draw a diagram : you want to carry some structure of $G$ (an automorphism $\alpha$) to $H$ and you're not sure how to compose the morphisms, then simply draw :
$\require{AMScd} \begin{CD} G @>{\phi}>> H;\\ @VVV^\alpha @VVV^\beta \\ G @>{\phi}>> H; \end{CD}$
($\alpha$ is supposed to be on the leftmost arrow, if someone can help me fix this...)
This tells you that an automorphism that's naturally associated to $\alpha$, call it $\beta$, would satisfy $\beta\circ \phi = \phi\circ \alpha$, so $\beta = \phi\circ\alpha\circ \phi^{-1}$ seems like a good candidate.
Of course this is not a proof that this works, nor is it the only way to make it work, but this is a way of finding ideas, and it comes in handy more than once. Now you simply have to check that $\alpha \mapsto \phi\circ \alpha\circ\phi^{-1}$ is an isomorphism $Aut(G)\to Aut(H)$ !