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I think the following is true, but I'm not sure how to prove it. Any help would be greatly appreciated.

Suppose $G$ is a transitive subgroup of $S_n$ and $C$ is the centralizer of $G$ in $S_n$. If $\sigma\in C$ has order $k$, then $\sigma$ is a product of $n/k$ disjoint $k$ cycles.

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  • $\begingroup$ It suffices (easy check) to show that if $\sigma \in C$ fixes any point, then $\sigma$ is trivial. Suppose $\sigma(x) = x$. Since $G$ is transitive, for any other point $y$, there exists a $g \in G$ such that $g(x) = y$. But now $$\sigma(y) = \sigma(g(x)) = \sigma g(x) = g \sigma(x) = g(x) = y.$$ Thus $\sigma$ fixes every point and so $\sigma = e$. $\endgroup$ – user670344 May 10 at 0:25
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This is true and well known. The property that $C$ has in your conclusion is usually just called "semiregular".

(See https://en.wikipedia.org/wiki/Group_action_(mathematics) or https://groupprops.subwiki.org/wiki/Semiregular_group_action)

So what you want is that "the centraliser of a transitive group is semiregular". Googling this phrase gives a bunch of hits.

For example, see Theorem 4.2A in the "Permutation Groups" book by Dixon and Mortimer.

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