Does this correctly interpret the "$\cong$" between groups as "$=$" between sets? I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.  

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;
$f$, $\bar f$ bijections;
$\theta$, $\bar \theta$ 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:

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
   I_n     &amp;   &amp;    I_n      \\
    \da{f}         &amp;           &amp;     \da{\bar f}          \\
   G &amp; \ras{\psi} &amp;    \overline G  \\
    \da{\theta}         &amp;           &amp;     \da{\bar \theta}          \\
   \operatorname{Sym}(G) &amp; \ras{\varphi^{(\psi)}} &amp;    \operatorname{Sym}(\overline G)  \\
    \da{\varphi^{(f)}}         &amp;           &amp;     \da{\varphi^{(\bar f)}}          \\
   S_n &amp;  &amp;    S_n  \\
\end{array}
$$


  Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?
No, those things need not be equal "for trivial reasons": you can take $G = G'$, and $\theta = \theta'$, and $\psi$ to be the identity (although it doesn't even factor into the equation you're asking about), but take $f \neq \bar f$. Then trivially $\theta(G) = \theta'(G')$, but the mappings $f, f'$ give different isomorphisms $\phi^{(f)}$ and $\phi^{(f')}$, so that you get different images in $S_n$. Concretely, you could take $G = C_4$ with generator $g$, and let $\theta$ be the Cayley embedding, and let $f(k) = g^k$, while $f'(1) = g, f'(2) = g^3, f'(3) = g^2, f'(4) = e$, so that $(1, 2, 3, 4)$ is in $(\phi^{(f)} \circ \theta)(G)$ but not in $(\phi^{(f')} \circ \theta')(G')$.

The problem with this is not that group isomorphism is the wrong notion, it's that you're asking the wrong question about it. It would work better with the following changes: suppose that $f, f'$ have the additional condition that $\psi \circ f = f'$, and that $\theta, \theta'$ are the Cayley embeddings for $G, G'$, then do the two mappings $I_n \to S_n$ given by
$$
\phi^{(f)} \circ \theta \circ f \qquad \text{and} \qquad \phi^{(f')} \circ \theta' \circ f'
$$
agree? And indeed the answer is that they do. For take $i, k \in I_n$, then
$$
\begin{align*}
\phi^{(f)}(\theta(f(i)))(k) &amp;= f^{-1}(\theta(f(i))(f(k)))\\
&amp;= f^{-1}(f(i)f(k))\\
&amp;= f^{-1}\psi^{-1}\psi(f(i)f(k))\\
&amp;= (f^{-1}\psi^{-1})(\psi(f(i))\psi(f(k)))\\
&amp;= f'^{-1}(f'(i)f'(k))\\
&amp;= f'^{-1}(\theta'(f'(i))(f'(k)))\\
&amp;= \phi^{(f')}(\theta'(f'(i)))(k).
\end{align*}
$$
You cannot meaningfully have a bijection between a set and a group.

You can have a bijection between a set and the underlying set of a group, but the underlying set does not in general fix the group up to isomorphism, you can have non-isomorphic groups of the same order (same number of group elements). The simplest example being the cyclic group of order $4$ versus the Klein four-group:


The set $\{0,1,2,3\}$ together with addition modulo $4$ (that is, add the numbers, and then if you get beyond $3$, subtract $4$ to get back into the set) gives, up to isomorphism, the cyclic group of order $4$.
The set $\{0,1,2,3\}$ together with bitwise xor (that is write the number as two-bit binary, form the two-bit string that has an $1$ in positions where the bits of the two operands differ and a $0$ where they don't, and interpret the result as number) is isomorphic to the Klein four-group.


Those groups are very different. For example, in the cyclic group, $1+1=2\ne 0$, but in the Klein four-group, $n+n=0$ for all $n$ (nowhere do the bits of an element differ from the bits of itself!).

So your scheme fails at the very first step, defining a bijection between a set and a group.