Consider the conjugation action of $S_n$ on an $n$ elements subset of itself.
How many actions are possible such that surjective property is maintained on the $n$ element subset of $S_n.$
Let the $n$ elements subset be denoted as $X_n.$
Seems two equivalent approaches:
left application of conjugate action on $X_n,$
Or, equivalently right application of conjugate action on $X_n.$
Let, $\tau$ be subset of $S_n$ and gets as result of its action on $X_n,$ get permutation $\sigma.$
Taking first interpretation,
$\sigma = \tau(abc)\tau^{-1}.$
But, seems confusing as $\tau$ is unknown.
Next, second interpretation.
$\sigma = (bac)\tau(abc).$
But, even that doesn't help as nothing else is known about $\sigma, \tau.$
Edit: There will be three fixed points, in composition of three permutations: $\tau(abc)\tau^{-1}.$
Taking a particular example of $S_6$ and $a,b,c= 1,2,3:$
Let $$\tau=\begin{pmatrix} 4&1& 5& 3&2&6\\ 1& 5& 3& 2& 6& 4\\ \end{pmatrix}$$ then, $$\tau^{-1}=\begin{pmatrix} 1&5& 3& 2&6&4\\ 4& 1& 5& 3& 2& 6\\ \end{pmatrix}.$$ Then $$(123)\tau^{-1}= \begin{pmatrix} 1&5& 3& 2&6&4\\ 4& 2& 5& 1& 3& 6\\ \end{pmatrix},$$ Then, $$\tau(123)\tau^{-1}=\begin{pmatrix} 1&5& 3& 2&6&4\\ 1& 6& 3& 5& 2& 4\\ \end{pmatrix},$$
But, how to prove there will always be three fixed points, and the rest are in a cycle?
The fixed points in $\tau(123)\tau^{-1}$ are all those elements in $\tau^{-1}$ not having map (co-domain) in $\{1,2,3\}$; as the map is changed for others in $(123)\tau^{-1}$ to not yield fixed point after composition from left by $\tau$.
Conjugate elements of $S_n$ have the same cycle type. Conversely, elements with the same cycle type are conjugate.
So it's a matter of counting the number of $3$ cycles $\sigma $.
In $S_6$, there's $6\cdot 5\cdot 4/3=40$ three cycles.