Let's say we have a system of interacting particles that can divided into two populations. The symmetry group of each population is $G$, and the two populations are identical, so that I can exchange them and the system remains invariant (see figure). I think the symmetry under exchange of the two populations should be described by the symmetric group $\mathfrak{S}_2$, see here.
My question is: what is the symmetry group of the whole system? Intuitively it should be some kind of composition between $\mathfrak{S}_2$ and $G$, but which one? Thanks in advance for your help!
Let us formalize it the following way :
$$Y=X_1\cup X_2 $$
This is a disjoint union. Set $S(*)$ to be the symmetry of $*$. We know that $S(X_1)=S(X_2)=G$. Then $S(Y)$ acts on $Y$, set $S_1$ the set of symmetry fixing $X_1$ then they fix $X_2$, hence $S_1$ is $S(X_1)\times S(X_2)=G^2$. Now we have that $S(Y)/S_1$ is the group of permutation of $\{X_1,X_2\}$ hence it is $\mathfrak{S}_2=\mathbb{Z}/2\mathbb{Z}$. Now I claim that the quotient admits a section hence you have :
$$S(Y)=(G\times G)\rtimes_{\phi} \mathfrak{S}_2$$
Finally we have to determine $\phi$, I claim that $\phi((1,2)).(g_1,g_2)=(g_2,g_1)$ so that it is non trivial and it just permuts both coordinates. So you have your group.
Furthermore the same technique can be used to show that if $Y=X_1\cup...\cup X_n$ then :
$$S(Y)=G^n\rtimes_{\phi}\mathfrak{S}_n $$
Where $\phi(\sigma).(g_1,...,g_n)=(g_{\sigma(1)},...,g_{\sigma(n)})$. We actually say that this is the wreath product of $G$ with $\mathfrak{S}_n$ (see https://en.wikipedia.org/wiki/Wreath_product) :
$$S(Y)=G\wr \mathfrak{S}_n $$
Edit : the fact that $S(Y)/S_1$ admits a section obviously depends on your situation, but it most certainly has. Anyway the wreath product is usually what you need to formalize this situation : $n$ objects with each the same symmetries ($G\times...\times G=G^n$) and the symmetries of their configuration ($S(Y)/G^n$).