Show up to equivalence that there is only one group action of $\mathbb{Z}_{3}$ on $X=\left \{ 1,2,3 \right \}$. 
Show up to equivalence that there is only one group action of $\mathbb{Z}_{3}$ on $X=\left \{ 1,2,3 \right \}$.

I am not sure where to begin with such a proof. I would assume it is proof by contradiction and I use the definition of equivalent actions? Is there an easier way, because I am not very knowledgable on equivalent actions and how to apply this to the proof! Any help and guidance is appreciated.
 A: Remember that $\mathbb{Z}_3$ is cyclic generated by some element g. So any action on X is determined by what g$\cdot$X is, with the added condition that g$^3$ = $e$. Also, I think you mean there's one non-trivial action.
To show that two actions $\phi_1,\phi_2$ are equivalent, you will have to show  that there is some bijection f:X$\rightarrow$X such that $\phi_1$(g,x)=$\phi_2$(g,f(x)) $\forall g \in G$. 
A: This is false although close to truth. There are two possible group actions on $\{1,2,3\}$.
By $G$-set I will understand $(X,\cdot)$ pair where $X$ is a set and $\cdot:G\times X\to X$ is a group action.
First of all note that any orbit is isomorphic to $G/H$ as a $G$-set for some subgroup $H$. Since $G=\mathbb{Z}_3$ has only two possible subgroups, namely $G$ and $0$ then there are only two possible orbits: $G/0$ and $G/G$. Non-isomorphic because of different cardinality.
Now since every $G$-set is a disjoint union of orbits (up to $G$-isomorphism) than $X=\{1,2,3\}$ has two possible $G$-set structures: namely $X\simeq G/0$ or the trivial one $X\simeq G/G\sqcup G/G\sqcup G/G$.
