This question is derived (as want to derive the formula for below problem, and also general approach) from :
Compute the number of distinct actions of $C_m$ on set $X,$ s.t. $|X|= n= 2m+1.$
Let there be $i= n/m$ disjoint cycles, of length $m$, in a permutation of length $n$.
Then, the ways to obtain such permutations is given by:
$\frac{((nCm)(n-mCm)...(n-imCm))}{i}*2^i.$
Where the denominator is for ordering among the disjoint cycles.
While, the multiplier ($2^i$) is for two possible orderings in each.
If $n= 7,$ then have products of the form $(abc)(def),$ and the number of product of $3$-cycles given by:
$(((7C3)(4C3))/2) * 4 = $ $((7.5).2).4)= 35.8= 280.$
In case want to generalise for different values of $m,$ then need to find multiplier factor accordingly.
Say, if $m=4,$ have: $i=n/4$, multiplier is given by number of different ordering possible: $1234, 1243, 1324, 1342, 1423, 1432 = 4C2= 6.$
$((((7C4)/6).(3C3))/2) * 4 = $ $((7.2.5)= 70.$ Where divisor of $2$ is for orderings among the two cycles.
Similarly, want to find for $m=5,$ have: $i=n/5$, multiplier should be given by number of different ordering possible given by $5C2= 10$ if a general formula to find denominator were possible. But, instead, it is given by $18$ as given below:
$12345, 12354, 12435, 12453, 12534, 12543,$
$13245, 13254, 13452, 13425, 13524, 13542,$
$14235, 14253, 14325, 14352, 14523, 14532, $
Edit: precursor to this post.
Remember that an action is a homomorphism from $C_3$ into $S_7$, here. (A group action on a set $X$ always gives a homomorphism from $G$ into $\rm{Sym}(X)$.)
The $3$ inequivalent actions are $1\to e,\,1\to(123)$, and $1\to(123)(456)$, for instance.
There are $7\cdot 6\cdot 5/3=70$ three cycles.
And there are $70\cdot 4\cdot 3\cdot 2/(3\cdot 2)=280$ products of disjoint three cycles.
So $350$ elements of order three.
$1$ in $C_3$ can go to any of these, or to $e$. That's $351$ actions (total).