For a given group $G$ , what are the sets on which a non-trivial group action of $G$ can be defined ? Say we are given a group $G$ , we want to find those sets on which we can define an action of $G$ ; now in this sense any set $X$ works as we can always define the trivial action $o:G \times X \to X$ by $gox:=x$ ; but what if we want to allow only non-trivial actions ? that is for a given group $G$ can we always define a non-trivial group action on any set $X$ ? If not then can we classify those sets , depending on $G$ , on which a non-trivial group action of $G$ can be defined ? 
 A: Every group action of $G$ is uniquely a disjoint union of transitive group actions. These are all of the form $G/H$ where $H$ is a subgroup of $G$, and two conjugate subgroups give isomorphic group actions. So the problem reduces to classifying the conjugacy classes of subgroups of $G$, which is hard in general, but that's the general answer. The resulting action is nontrivial as long as at least one $H$ appears which is not all of $G$. 
A: All that matters is the cardinality of $X$. If $G$ acts nontrivially on $X$ and $X \subset Y$, then $G$ can act nontrivially on $Y$ by setting $g \cdot y = y$ if $y \in Y \setminus X$ (and otherwise it's the action on $X$ already defined). In the end the question becomes:

What's the smallest cardinal on which $G$ can act nontrivially?

And then $G$ can act nontrivially on all sets of bigger cardinality.
It's not always possible to define a nontrivial action of $G$ on a given set $X$ (even when $|X| > 1$). For example $\mathbb{Z}/3\mathbb{Z}$ cannot act nontrivially on a set with two elements (there are no nontrivial morphisms $\mathbb{Z}/3\mathbb{Z} \to \mathfrak{S}_2 = \mathbb{Z}/2\mathbb{Z}$).
In general I don't think the answer is known. If $\varphi : G \to \mathfrak{S}$ is a nontrivial morphism, then $G / \operatorname{ker}(\varphi) \cong \operatorname{im}(\varphi) < \mathfrak{S}$, so a quotient of $G$ is isomorphic to a subgroup of $\mathfrak{S}$. So if for example $G$ is simple of order $k$, then $G$ cannot act nontrivially on any set $X$ of cardinality $n$ with $n! < k$. With the existence of infinite simple groups the answer appears to be even more out of reach. In general the answer is "the smallest cardinality such that a quotient of $G$ embeds in the symmetric group", but that's not really tractable.
Some bounds: $G$ always acts nontrivially on itself, so $|G|! = |\operatorname{Bij}(G)|$ is an upper bound. Sometimes you cannot do better, eg. when $G$ is simple.
