Every group action on a set $S$ partitions the set into orbits. Conversely, for every partition of $S$ is there a group action such that the set of orbits of the group action equals the partition?
My attempt. Let $P = \{S_1,\dots, S_k\}$ be a partition of a finite set $S$ of order $n$. Then there are elements of $G = Perm(S)$, the set of permutations of $S$, that leave each subset fixed, i.e. $g\in G$ such that $g\cdot S_i = \{gs : s\in S_i\} = S_i, \forall i$. Think of the permutations that fix all of $S$ except permutes $S_i$ possibly in some nontrivial way. If $g, h$ are such elements then $gS_i = S_i, \forall i$, so $g^{-1} S_i = S_i$, the action also being an action on the subsets. Similarly $g\cdot h S_i = g S_i = S_i$. Thus, the set of all $P$-stabilizing elements is a subgroup of $G$.
Groups are known in correspondence with subgroups of $G$. Please comment.
Motivation: is there a group-theoretical way to count partitions of a set of size $n$? So that maybe it can help prove formulas about the latter.
According to m_l in a comment, there's no application here to counting partitions here that doesn't lead to having to count the partitions the known ways.
I was thinking, we haven't considered yet, at least I haven't, the set of partitions of $S$ itself, call it $P(S)$, and the group $G = Perm(S)$ acting on it. The nice thing about $G = Perm(S)$ is that every finite group is isomorphic to a subgroup of it. So if $H \leqslant G$, and $H'$ is the group presented in some other way, then almost anything we say about $H$ can be applied to $H'$ regarding group actions on $S$ or $P(S)$. Hopefully, so let's keep that in the back of our minds. In other words, every partition corresponds to a partition of the set of subgroups of $G$. But we should say more about it than that. I'm moving off track here, so continuing on…
Define the action of $G$ on $P(S), $ to be $g \cdot P = \cup_{i=|P|} \{ g\cdot S_i \}$, where $g\cdot S_i$ is the element $g$ acting on the block $S_i$ suing simple coset multpilcation. Relatedly there's an obvious way to make $G$ act on the set of all subsets of $S$.
Anyway. As you can tell the action on $P(S)$ defines a group action. Proof:
The identity permutation obviously fixes a partition. And $g(h P) = g\cdot\cup_{i=1..|P|} \{ h\cdot S_i \}$. Note that since $g$ acts on the set of subsets of $S$, $g(hP) = (gh)P. \ $ The rest of the proof is left to the reader.
Let's switch notation a bit. $P\in P(S)$ will now be called $p$, and $P(S)$ will be called $P$.
An orbit is simply $O_p = G\cdot p \subset P(S)$. We have the class equation
$|P(S)| = \sum_{orbits} |O_p|$
Burnside's Lemma:
# orbits $ = 1/|G| \sum_{g\in G}|P^g|, \ $ where $P^g$ is the set of all partitions fixed by $g$.
and the index counting formula
$|G| = |H_p||O_p| = |H_p|[G:H_p], \ $ where $H_p = Stab(p)$ is the stabilizer of the partition $p$.
Best Answer
Let $S$ be a set (not necessarily finite) and $\mathcal{P}=\{P_i\}_{i\in I}$ some partition of $S$. Define
$$G_i:=\{f:P_i\to P_i\ |\ f\mbox{ is invertible}\}$$
With standard function composition $G_i$ is a group. If $P_i$ is finite then it is the standard symmetric group $S_{|P_i|}$. It acts on $P_i$ via $(f, x)\mapsto f(x)$. You can easily check that $P_i$ is transitive (i.e. it has only one orbit) under this action.
Now put $G:=\prod G_i$ and define the action
$$\big((f_i), x)\big)\mapsto f_j(x)\mbox{ if }x\in P_j$$
You can easily check that $S/G=\mathcal{P}$.
Side note: If we assume the Axiom of Choice or assume that each $P_i$ is at most countable then we can make $G$ smaller. Just define $G_i:=P_i$ with any group structure and put again $G:=\prod G_i$. Note that the statement "any set can be turned into a group" is equivalent to the Axiom of Choice. However it is true for at most countable sets independent on the Axiom of Choice.
Side note2: The $G$ can be refined even more if each $P_i$ is finite and $I$ is finite. Define $G:=\mathbb{Z}_n$ where $n=lcm(|P_i|)$ (least common multiple). Since $P_i$ is equinumerous with $G/H$ for some $H$ then it can be easily equipped with transitive action. That's probably the smallest $G$ we can get for finite case.
Side note3:
No. That's not a nice thing. It actually is an epicly horrible thing, because it means that proving anything about $G$ is at least as hard as proving anything about any other group.