Question about symmetric group action I am trying to analyse the following.

Assume $S_{3}$ acts on a non empty set $T$, and that is has $3$ orbits. What can we say about the possible cardinalities for the set T? 

My thoughts:
If $G=S_{3}$ then $|G|=3!=6$ and we are consider $$G \times T \to T,~~(g,t) \to g \star t$$
I also know that $|\text{orb(t)}|=\frac{|G|}{|\text{stab(t)}|}$ 
I know lagranges thereom tells us the size of a subgroup must divide the size of the group. I also know that orbits partition a finite set, so then since orbit is a subgroup of T, we know the size of the orbits must divide it?
$$|T|=|orb(t_{1})|+|orb(t_{2})|+|orb(t_{3})|$$
And also, any fixed point is also a stabilizer. But I am having trouble putting all this together to make a coherent thought of what this tells me about T. 
Else I am just not so sure how to proceed.
Any advice? Thanks!

When you say "I also know that $\lvert \text{orb}(t)\rvert=\frac{\lvert G\rvert}{\lvert \text{stab}(t)\lvert}$ which tells us $\lvert \text{stab}(t)\rvert =2$," I think you are mistaking $\lvert \text{orb}(t)\rvert$ for the number of orbits of $T$. This is not what $\lvert\text{orb}(t)\rvert$ is. Given a particular element, $t\in T$, $\text{orb}(t)$ is the set of elements that to which the action of $G$ carries $t$, and $\lvert\text{orb}(t)\rvert$ is the number of these elements, and does not say anything about the other orbits of $T$.
Clearing this up may help you on your way to solving, the problem, but if you still need a hint knowing that $T$ has three orbits, we can pick elements $t_1, t_2, t_3\in T$ so that $$\lvert T\rvert=\lvert\text{orb}(t_1)\rvert+\lvert\text{orb}(t_2)\rvert+\lvert\text{orb}(t_3)\rvert\text{.}$$ 
