Is the following true?
Suppose $G$ is a transitive subgroup of $S_n$ and $G_1$ is the point stabilizer of 1 in $G$. Let $N$ denote the normalizer of $G_1$ in $G$. Suppose $\sigma\in N\setminus G_1$ has order $k$. Then $\sigma$ is a product of $n/k$ disjoint $k$ cycles; that is, the cyclic subgroup generated by $\sigma$ has $n/k$ orbits, all with $k$ elements.
That really seems very unlikely, and it is not difficult to find counterexamples. There is no reason why $k$ should divide $n$.
Try $S_5$ acting on the cosets of $A_4$, giving a subgroup of $S_{10}$. Then $N \cong S_4$, and elements of order $4$ have orbits of lengths $2$ and $4$.