abstract algebra - If $G to G'$ are isomorphic groups, then $|Aut(G)| = |$ all isomophisms $G to G'|$.

I'm having trouble proving this.

If $G$ is finite (which is not given) with $|G| = N$, then $Aut(G)$ can be seen as a symmetry group with order $N!$, so there are $N!$ possible isomorphisms $G \to G'$. So intuitively I understand why $|Aut(G)| = |$all isomorphisms $G \to G'|$.

My first thought was, try to find a bijection between $Aut(G)$ and $H = \{$all isomorphism $G \to G'\}$, by using a fixed $f: G \to G'$. But now I'm stuck.

I'm not looking for a proof, but rather some more insight/help with the thought process.