Let $G$ be any non-cyclic finite group generated by two elements $g$ and $h$. Take the regular action of $G$ on itself, i.e., let $G$ act on itself by right rsp. left multiplication. Identify $G$ (as set acted upon) with a subset of $\mathbb N$ where the $1$ of the group is identified with $1 \in \mathbb N$, $g$ is identified with $2$ and $h$ with $3$.
As the orbits of the cyclic subgroups of $G$ containing $1$ are just the cyclic subgroups, none of them containing both $g$ and $h$, $G$ considered as subgroup (the regular action is faithful) of $\cal F$ is minimal singular.