Quaternion group $Q_8$ has though $6$ elements of order $4$, but there are three sets ($(i, -i), (j, -j), (k, -k)$) having seperate subgroups, with identity and $-1$ common only.
Apart from that have one element of order $2$, and $e$.
But, how to use that information, is unclear.
It is not a cyclic group, hence cannot simply state based on orders of elements.
Best Answer
Hint Just like in your previous post about transitive actions of $C_{12}$, the general result (that @Derek Holt clued us in on) is that there are as many inequivalent actions as conjugacy classes of subgroups (of, in this case, $Q_8$).
The aforementioned also gave us a headstart in the comments that in this case there are $6$ conjugacy classes of subgroups.
The subgroups are: $, \{1\},\langle-1\rangle, \langle i\rangle, \langle j\rangle, \langle k\rangle $ and $Q_8$.
Automorphisms preserve the order of subgroups, so the only question is if the $3$ subgroups of order $4$ are conjugate. Since they each have index $2$, they're all normal. So not conjugate. That's where the number $6$ comes from (each subgroup is its own conjugacy class).
Here's some terminology: the quaternions are Hamiltonian. That's they're a non-abelian group such that every subgroup is normal.
In this problem, that tells us the number of transitive actions is equal to the number of subgroups.