Study of the number of the orbits of $S_4$ acting in $X$ with $|X|=36$

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The objective is to check some statements and these are questions that I was unable to answer:

Consider the number of the orbits of $S_4$ acting in $X$ with $|X|=36$

i) Can we have exactly $15$ orbits?

ii) The number of orbits must divide $24$?

iii) Can we have only one orbit?

I tried to apply the Burnside's Lemma, since it involves the number of orbits in the action. But I really don't know how I can get the results.

For the first question I see that $\sum |X^g|=15\cdot24$, but how do I know this is impossible ?

For the second question I don't see why it should divide $24$, but I'm not sure I can say that the answer is no.

For the third one, if I have only one orbit I suppose that $\forall x,y \in X\,\,\ \exists g \in S_4$ such that $g\cdot x = y$, but how can I use that (if true) ?

I'm a begginer in Group Theory and I imagine these are easy questions, but I can't see how to answer then. I'm sorry in advance for any English or Math errors.

Thank you.

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Sketch solution: $|S_4|=4!=4\cdot3\cdot2=24$ and we know that it acts on $X$, whose number of elements is $36$.

The order of a generic orbit of an element $x\in X$ is such that $$|\mathcal O(x)|=|S_4:Stab(x)|\implies|\mathcal O(x)|\big|24.$$ This means that $|\mathcal O(x)|\in\{1,2,3,4,6,8,12,24\}$ and the action induces a partition on the set $X$.
A possible partition, in order to have $15$ orbits, could be $(2,2,2,2,2,2,2,2,2,2,2,2,2,2,6)$.
The action defines an equivalence relation in $X$ and the $\textbf{order}$ of each orbit (which are equivalence classes respect to the relation $a\sim b$ if $\exists\sigma\in S_4:b=\sigma*a$) divides the order of the group $S_4$.

The number of the orbits is one if the action is transitive, so, as you wrote, if $\forall a,b\in X$ $\exists\sigma\in S_4$such that $b=\sigma*a$ and in this case there is only one equivalence class (the partition is $(36)$).
Since the order of an orbit is a divisor of $|S_4|$, we can't have only one orbit.