Let $S_{n}$ be the permutation group of $n$ elements. If $F$ is finite group, then a homomorphism $f : F \rightarrow S_{n}$ is called regular if the corresponding action of $F$ on $E$, the set of $n$ elements, is free, i.e. if $f(x).e=e$ for some $x\in F$ and $e\in E$, then $x=1$.
My question is that is it true that if the order of group $F$ divides $n$ then there is a regular homomorphism $f : F \rightarrow S_{n}$ and such a homomorphism is unique up to conjugation by an element of $S_{n}$?
Write $n=pq$ where the order of $F$ is $p$ consider the set o $np$ elements $F^q$ on which $F$ acts b left translations.