[This post is admittedly a development of this other of mine, based on therein accepted answer.]
Here I kindly ask for a "solution verification" of my final interpretation.
For $n$ positive integer, let be:
- $I_n:=\{1,\dots,n\}$;
- $G$, $\overline G$ groups of order $n$;
- $\psi\colon G \rightarrow \overline G$ isomorphism;
- $f$ bijection and $\bar f = \psi f$;
- $\theta$, $\bar \theta$ Cayley embeddings;
- in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets;
- $S_n$ the symmetric group of degree $n$.
Visually:
Then:
$$\varphi^{(f)} \theta f = \varphi^{(\bar f)} \bar \theta \bar f \tag 1$$
Proof. Since $\varphi^{(\bar f)}=\varphi^{(f)}(\varphi^{(\psi)})^{-1}$ and (*) $\bar\theta=\varphi^{(\psi)}\theta\psi^{-1}$, we get:
$$\varphi^{(\bar f)} \bar\theta \bar f = \varphi^{(f)}(\varphi^{(\psi)})^{-1}\varphi^{(\psi)}\theta\psi^{-1}\psi f = \varphi^{(f)}\theta f$$
$\Box$
My interpretation of $(1)$ is as follows:
If two (finite) groups are isomorphic (via $\psi$), then, for a given labelling of the elements of the first one (via $f^{-1}$), a labelling of the elements of the second one exists (given by $f^{-1}\psi^{-1}$) such that the two groups' structures manifest in $S_n$ as one same set (by $(1)$, actually).
Edit. As pointed out in the comments, (*) needs not to be true, unless the embeddings are Cayley's. Anyway, since these latter always exist, the conclusion keeps unaltered.
Edit #2. If we'd define structure of $G$ (under a given labelling $f^{-1}$) the map $\varphi^{(f)}\theta f$, where $\theta$ is Cayley embedding of $G$ into $S_G$, then $(1)$ would precisely state that isomorphic groups have the same structure.
Edit #3. I think we can upgrade this result to an "iff" one.
Let:
where, for $n$ positive integer:
- $I_n:=\{1,\dots,n\}$;
- $G$, $\overline G$ are groups of order $n$;
- $f$, $\bar f$ are bijections;
- $\theta$, $\bar \theta$ are Cayley embeddings;
- in general, $\varphi^{(\alpha)}$ is the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets;
- $S_n$ is the symmetric group of degree $n$.
Claim. Let $f$, $\bar f$, $G$, $\overline G$, $\theta$ and $\bar \theta$ be as above. Then $G \cong \overline G$ if and only if:
$$\exists f,\bar f \mid \varphi^{(f)}\theta f = \varphi^{(\bar f)}\bar \theta\bar f \tag 1$$
Proof. $\Rightarrow$) Let $\psi$ be the isomorphism between $G$ and $\overline G$; then, for any $f$, the bijection $\bar f := \psi f$ does the job (see here for the proof). $\Leftarrow$) Let $(1)$ hold for some $f$ and $\bar f$; then $\bar f f^{-1}$ is an isomorphism from $G$ to $\overline G$ (see here for the proof). $\quad \Box$
Yes, your interpretation is correct.
It means that corresponding elements (of $G$ and $\bar G$) receive the same label (from $I_n$ where $n=|G|$), and by the isomorphism they behave exactly the same way (generating the exact same permutations on the labels).