= Solution
A <cyclic group> of order six has exactly one <subgroup> of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of <subgroups>. Thus there is one transitive <group action> of each size $1,2,3,6$.
An action on six points is a disjoint union of its <orbits>. Its equivalence type is therefore determined by nonnegative integers satisfying
$$
n_1+2n_2+3n_3+6n_6=6.
$$
If $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. Hence
$$
\boxed{1+4+2+1=8\text{ inequivalent actions}.}
$$
This is <counting finite group actions by transitive orbit types>. Equivalently, a generator acts as a <permutation> whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.
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