Let $G$ be a group, $S$ be a set, $H \trianglelefteq G$, $a,b \in S$, and suppose $G$ acts faithfully and transitively on $S$. If $H \leq G$, then prove $|Ha|=|Hb|$ for every $a,b \in S$. In short, we are to prove that all of the orbits $H$ has are of the same size.
First, since $G$ acts on $S$ and since $H$ is a subgroup of $G$, then $H$ will also act on $S$ by the same action. First, we know for any $a \in S$, we have
$$|Ha| = \frac{|H|}{|H_a|}$$
where $H_a$ is the stabilizer of $a$ in $H$. Then, to show that $|Ha| = |Hb|$ for every $a,b \in S$, we really must show that
$$\frac{|H|}{|H_a|} = \frac{|H|}{|H_b|} \implies |H_a| = |H_b|$$
Since $G$ is faithful, if $g\in G$ with $ga = a$, then $g=1$. So $G_a =\{1\}$ and so $|G_a| = 1$. Since $H \leq G$, it must be true that $|H_a| =1$ and $|H_b|=1$. But is this all? Or have I missed something? This seems too easy to be true. For example, transitivity wasn't used here.
UPDATE: I am now realizing that a faithful action does not imply that $G_a = \{1\}$. This ruins the entire proof. How can I use faithfulness and transitivity to show what is needed?
UPDATE 2: I have stipulated now that $H \trianglelefteq G$
UPDATE 3: I will ask a new question elsewhere.
This is not true. Without loss of generality, we might as well take $G$ to be the symmetric group on $S$. This is faithful and transitive, and every permutation group on $S$ is a subgroup of $G$, so now your claim is simply that every permutation group has all orbits of the same size. It is easy to find many counter-examples, for example the group generated by a transposition on a set of size bigger than two, or the point-stabiliser $S_{n-1}$ of $S_n$ (again with $n\geq 3$).