Can $G≅H$ and $G≇H$ in two different views? Can $G≅H$ and $G≇H$ in two different views?
We have two isomorphic groups $G$ and $H$, so $G≅H$ as groups and suppose that they act on a same finite set, say $\Omega$. Can we see $G≇H$ as permutation groups. Honestly, I am intrested in this point in the following link. It is started by:

Notice that different permutation groups may well be isomorphic as ....

See here
 A: The definition of "isomorphic as permutation groups" given in the wikipedia page you refer to is equivalent to the images of $G$ and $H$ in their actions on $\Omega$ being conjugate in the symmetric group on $\Omega$.
You should be able to think of examples of subgroups $G$ and $H$ of $S_n$ for some $n$ such that $G$ and $H$ are isomorphic as groups, but not conjugate in $S_n$. Try groups of order 2 in $S_4$, for example.
A: Let $G$ and $H$ act faithfully on a finite set $\Omega$. This is equivalent to say that $G$ and $H$ are subgroups of the total permutation group $\mathfrak S_\Omega$.
Here are two notion of isomorphism which are stronger than a simple group isomorphism :


*

*The action of $G$ and $H$ are isomorphic if there exists a group morphism $\phi : G \to H$ and a bijection $\sigma : \Omega \to \Omega$ such that for all $g\in G$ and $x\in \Omega$
$$ g\cdot \sigma x = \sigma(\phi g \cdot x). $$
In particular, the morphism $\phi$ is an isomorphism, because if $\phi g = 1$ then for all $x\in\Omega$ $g\cdot \sigma x = \sigma x$, and since the action of $G$ is faithful and $\sigma$ surjective, this implies that $g = 1$.

*$G$ and $H$ are said to be conjugate in $\mathfrak S_\Omega$ is there exist a permutation $\sigma$ of $\Omega$ such that $G = \sigma H \sigma^{-1}$. In particular the map $g\in G \mapsto \sigma^{-1} g \sigma\in H$ is an isomorphism.
In fact, both notion are the same (easy exercise for you !), and its what Wikipedia call permutation group isomorphism.
This notion is strictly stronger that the notion of group isomorphism. For example, take $\Omega = \{1,2,3,4\}$, $G$ the group of order $2$ generated by the transposition $(1 2)$ and $H$ the group of order $2$ generated by the double transposition $(1 2)(3 4)$. As groups, $G$ and $H$ are isomorphic, because they are both isomorphic to $\Bbb Z/2\Bbb Z$. However, they are not isomorphic as permutation group. Indeed, the conjugate of a transposition is always a transposition, it cannot be a double transposition. More precisely, the conjugate $\sigma (1 2) \sigma^{-1}$ is the transposition $(\sigma 1, \sigma 2)$.
When classifying the subgroups of a given group, it is often important to classify them up to isomorphism but also up to conjugation, because isomorphism class can split into several conjugation classes.
