Counting number of isomorphisms between graphs using its automorphism group

Question

Define $\mathcal{G}_n$ to be the set of labeled graphs whose vertices are $v_1,v_2,..,v_n$.
Let $G,H \in \mathcal{G}_n$ such that $G \cong H$ (They are isomorphic), let $\theta$ be the isomorphism.
I need to show that the set of all isomorphisms between G to H is exactly the coset $\theta \operatorname{Aut}(G)$, where $\operatorname{Aut}(G)$ is the automorphism group of G. This is what I've done so far:

Let $\operatorname{Iso}_G ^H$ be the set of isomorphisms from $G$ to $H$, I need to prove that $\operatorname{Iso}_G ^H=\theta \operatorname{Aut}(G)$.
I've already shown that for every $\alpha \in \operatorname{Aut}(G)$, the composition $\theta \alpha$ is a isomorpishm, so $\operatorname{Iso}_G ^H \supseteq \theta \operatorname{Aut}(G)$ is immediate.
For the other direction, let $\theta_0 \in \operatorname{Iso}_G ^H$. If $\theta_0=\theta$ pick the identity permutation from $\operatorname{Aut}(G)$. (Which always contains atleast it)

Now this is the part where I'm stuck.

I know that two different isomorphisms agree on some vertices $v,u$ (meaning $\theta_0(v)=\theta(u)$) $iff$ $v$ and $u$ are similar. (There's a automorphism that maps them to each other)
So in a hand-wavy fashion - since similarity is an equivalence relation, I can always find $\alpha \in \operatorname{Aut}(G)$ that "repairs" the permutation of specific equivalence class such that $\theta$ will act will act on them in same way that $\theta_0$ does. How can I capture that intuition in more rigorous manner? Or if my intuition is wrong, how can I prove that $\operatorname{Iso}_G ^H \subseteq \theta \operatorname{Aut}(G)$?

Answer 1

In general, if $X$ and $Y$ are two isomorphic objects (graphs, sets, groups, etc...) and $\theta : X \to Y$ is a fixed isomorphism, then $$ \begin{align*} \gamma \mapsto \theta \circ \gamma: \mathrm{Aut}(X) \to \mathrm{Isom}(X,Y) && && (*) \end{align*}$$ is a bijection. Since compositions of isomorphisms are isomorphisms, we know in particular that $(*)$ indeed defines a mapping $\mathrm{Aut}(X) \to \mathrm{Isom}(X,Y)$.

But why is $(*)$ a bijection? Well, the point is that we can write down the inverse mapping. $$ \begin{align*} \gamma \mapsto \theta^{-1} \circ \gamma : \mathrm{Isom}(X,Y) \to \mathrm{Aut}(X) && && (**) \end{align*}$$ is a map in the other direction, and you can easily check that $(*)$ followed by $(**)$ is the identity, and similarly $(**)$ followed by $(*)$ is the identity.