$\Sigma\le S_n$ of order $n$. Is $\Sigma$ cyclic, if each $\tau\in\Sigma\setminus\{Id\}$ moves all the elements?

Question

If $\sigma\in S_n$ is an $n$-cycle, then every $\tau\in\Sigma:=\langle\sigma \rangle$ moves all the elements of $\{1,\dots,n\}$. Now I wonder whether the other way around holds, namely:

Claim. Let $\Sigma$ be a subgroup of order $n$ of $S_n$, such that every $\tau\in\Sigma\setminus\{Id\}$ moves all the elements of $\{1,\dots,n\}$. Then, $\Sigma$ is cyclic.

Is the claim true?

My thoughts: by contrapositive, if $\Sigma$ is not cyclic, then it is generated by more than one permutation; I'm thinking that some of them must fix some element (?), so $\Sigma$ hasn't got the desired property.

Answer 1

$S_4$ contains Klein's group in the form $\{e, (12)(34), (13)(24), (14)(23)\}$. The group does act freely on $\{1,2,3,4\}$, but it is not cyclic.

Answer 2

Hint

Actually, for a counterexample just note that any group embeds in $S_{\lvert G\rvert}.$

And it does so via derangements. That's the content of Cayley's theorem. The action of $G$ on itself is both free and faithful.

So just choose any noncyclic group, and it's isomorphic to a permutation group with the desired property.