If G is a group and it acts on X and it has 6 orbits.H is a subgroup of G, with the property that $[G:H]=3$ (H has only 3 left cosets), how many orbits can it have in X?
By using the orbit-stabilizer theorem: $|G|=|Gx||G_x|$ and so $|H|=\frac{1}{3}|Gx||G_x|$ but I do not know how to use this information..
This is not an answer but an example that might point you in the right direction.
Let $G=\mathbb{Z}_6$ and $X=\left\{re^{\frac{ k 2\pi i}{6}}\mid r,k\in \left\{1,2,3,4,5,6\right\}\right\}$. Let $\overline{1}$ in $G$ act on $X$ by rotation over $60$ degrees with respect to the origin, i.e. $\overline{1}\cdot re^{\frac{k2\pi i}{n}}= re^{\frac{(k+1)2\pi i}{n}}$. It is easy to check that this action has $6$ orbits.
Let $H=\left\{\overline{0},\overline{3}\right\}$, then $H$ is a subgroup of index three. The action restricted to $H$ is determined by the action of $\overline{3}$. Clearly, $\overline{3}$ acts as a flip (rotation over $180$ degrees). Each orbit now has $2$ elements in it, thus there are $18$ orbits. Can you guess a formula from this example?