Let $H$ be a transitive subgroup of $\mathfrak{S}_n$, $n \geq 2$. Using Jordan's lemma ($H$ is not a union of conjugate proper subgroups), we see that $H$ contains a permutation without fixed points. I'm interested in whether $H$ necessarily contains a permutation whose cycle lengths have a common divisor. It's tricky to either prove or disprove this: on the one hand, my attempts at proving this have all been vain (some cases are easy: if $H$ embeds via its action on itself, if $H$ is a $p$-group ; if $n$ is a prime power we can reduce to the $p$-group case by noticing that any $p$-Sylow of $H$ acts itself transitively); on the other hand, a colleague of mine has checked that this is true for $2 \leq n \leq 27$ using Julia (assuming correctness of his code). A key observation based on his experimentations is that you cannot choose which of the prime divisors of $n$ divides the cycle lengths of an element (there is a transitive subgroup of $\mathfrak{S}_6$ with no elements having only even-sized cycles, and a transitive subgroup of $\mathfrak{S}_{12}$ with no elements having only cycles of length a multiple of three, so neither the smallest nor the largest prime dividing $n$ are always candidates for example).

Thanks for any opinion about this question ☺