What is the maximal order $f(n)$ of a metacyclic (metacyclic group is the extension of a cyclic group by a cyclic group) transitive permutation group of degree $n$? It can be easily proved that $f(p)=p(p-1)$ for prime number $p$, and the following argument (if it is correct) shows that $f(n)$ divides $n^2\phi(n)$:

For a metacyclic transitive permutation group $R$ of degree $n$, we can write $R=\langle a,b\rangle$ with $\langle a\rangle\trianglelefteq R$. Let $M=\langle a\rangle=Z_m$ and $L=\langle b\rangle=Z_l$. Since the point stabilizer of a transitive permutation group is core-free, we have $R_\omega\cap M=1$ for any point $\omega$. This implies that $M$ is semiregular and hence $$ m~\big|~n. $$ Since the center of $R$, denoted by $Z$, is semiregular, we have $|Z|~\big|~n$. Then $C\cap L\leqslant Z$ and $L/(C\cap L)\lesssim Aut(M)=Z_{\phi(m)}$ where $C$ is the centralizer of $M$ in $R$ yields $$ l~\big|~n\phi(m). $$ Therefore, $ml~\big|~n^2\phi(n)$ and thus $|R|~\big|~n^2\phi(n)$.

However, computation results suggest that $f(n)\leqslant n(n-1)$ for all $n$ and $f(n)=n(n-1)$ only if $n$ is prime.