Call a number $n$ special in case there is a unique group of order n whose character table consists of only integers. This should be equivalent to the condition that the number of conjugacy classes coincides with the number of conjugacy classes of cyclic subgroups.

Questions: What is the sequence of special numbers? It starts with 1,2, 4, 6, 12, 18, 54,120. Is it an infinite sequence?

For which $m$ is $m!$ special ?(note that the symmetric group has integral character table and thus $n!$ is a good candidate for being special)

[Side question: For which m are the groups of order m! classified?]

What are the corresponding unique groups of order $n$ for $n$ being special?

The sequence of such groups starts with $\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2, S_3,D_{12}, (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2,(\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2, S_5$