Suppose, $G$ is a finitely generated group with a finite set of generators $A$. Suppose $G$ is acting on a set $S$. Let’s call such action a ping-pong action iff $\exists$ a collection of pairwise disjoint and non-empty subsets $\{S_a\}_{a \in A} \subset P(S)$ such that $\forall a \neq b \in A$ such that $\forall k \in \mathbb{Z}\setminus \{0\} a^k(S_b) \subset S_a$.
The following fact is well known about such actions:
If $G$ accepts a ping-pong action on some set, then it is freely generated by $A$.
A converse is also true, as the action of free group upon itself via left multiplication is a ping-pong action.
However, I wonder whether a stronger converse is true:
Suppose $F$ is a free group of finite rank faithfully acting upon a set $S$. Is such action necessarily a ping-pong action?
There's a problem with the definition, as I mentioned in the comment, but this can be fixed.
Once fixed: the answer is no:
indeed, there exists a faithful action of $F_2$ on a countable set, such that every orbit is finite (actually, a Baire-generic pair in the symmetric group $S_\omega$ yields such an action). Then there are no ping-pong subsets in any reasonable sense.