Is there a faithful transitive locally finite action of the modular group? Is there a faithful transitive action of $G = \mathrm{PSL}_2(\mathbb{Z})$ on $\mathbb{Z}$ such that orbits under each $g \in G$ are finite? 
 A: Say that a finitely generated group $G$ has property Z if it acts transitively on an infinite countable set $X$ such that the orbits of each element are finite.


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*If a finite index subgroup $H$ of $G$ has property Z, then $G$ has property Z. Namely, given $H$ acting on $X$, take the action of $G$ on the "induced permutation representation"


$$Y = \coprod_{G/H} gX$$
This is transitive, because the action of $G$ acts transitively on the factors, and then $H \simeq g H g^{-1}$ acts transitively on $gX$ by assumption. Because the action is transitive, to check the orbits of all $g \in G$ are finite, it suffices to check it for a single element $x \in X$. Also, to check that the orbit of any $g \in G$ is finite, it suffices to show that the orbit of some finite power of $g$ is finite. Yet, because $H$ has finite index, a finite power of $g$ lands in $H$, which then has a finite orbit on $x \in X$ by assumption.


*If a quotient $G/H$ has property Z, then $G$ has property $Z$. Explicitly, if $G/H$ acts on $X$, then there is a corresponding action of $G$ on $X$ which factors through $G/H$.


I claim that $G =\mathrm{PSL}_2(\mathbf{Z})$ has property $Z$. The group $G$ is finitely generated and virtually free, so, by (1), we may reduce to the case when $G$ is a finitely generated free group with at least any given number of generators. By (2), we may then reduce to the case of any finitely generated group we like. On the other hand, we may now take $G$ to be any finitely generated infinite (hence countable) group such that the order of any element $g \in G$ is finite, and to take $X = G$ with the tautological left action. Many such groups exist (Burnside groups, Grigorchuk group, Tarski Monsters...)
