Transitive subgroup of $S_n$ that has order n and consists of derangement.

Question

Besides $Z_n$, is there any subgroup of $S_n$ that has order n and the group actions are all derangements?

Answer 1

By Burnside's lemma, if a group $G$ acts transitively on a set $X$, then

$$|X/G| = 1 = \frac{1}{|G|} \sum_{g \in G} |\text{Fix}(g)|.$$

Since $|\text{Fix}(e)| = |X|$, if all the non-identity elements fix no points then we must have $1 = \frac{|X|}{|G|}$, so $|G| = |X|$, and this is necessary and sufficient; if $|G| = |X|$ and $G$ acts transitively on $X$ then by Burnside's lemma no non-identity element can have a fixed point.

The stabilizer of such an action must be trivial so $X$ must be isomorphic to $G$ being acted on by left multiplication. So every finite group of order $n$ occurs as a transitive permutation group on $n$ elements where every non-identity element is a derangement, and the smallest non-cyclic example is the Klein four group $C_2 \times C_2$ acting on itself by left multiplication; this embeds it as a subgroup of $S_4$ in the usual way.

Relaxing the condition slightly, transitive permutation groups where non-identity elements have at most one fixed point and at least one non-identity element fixes a point are Frobenius groups.