Let $A = (1 2 3 \dots n)$ be an element of $P_n$. So that the group $P_n$ acts on itself by means of the action of conjugation, for $B \in P_n$,
$$B · A = B A B^{-1}.$$
Stabilizer of $A$ is subgroup; $$\{A^{c} \mid c = 0, \dots , n-1\}.$$
I want to find the cardinality of $Orb_{P_n}(A)$.
So by applying the Orbit stabilizer theorem:
Then for A to act in itself by conjugation then I have that $Orb_{P_n}(A)=ABA^{-1}$ : where $A$ is an element of $P_n$.
[Provided by my friend Andreas]
Proof of the stabilzer of A:
By applying the orbit-stabilizer theorem, you can count how many n-cycles there are.
The result is $n! / n = (n-1)!$, as each $n$-cycle can be written in $n$ different ways, $$ (12\dots n) = (23\dots n 1) = \dots. $$
Since the conjugacy class of $(12\dots n)$ consists of all $n$-cycles, orbit-stabilizer tells you that the stabilizer you are looking for (which is called the centralizer of $A$) has order $n$. Since there are n distinct powers of A, and they clearly stabilize A, and so forth.
Is this correct :O
-nomad609
The centralizer of $A=( 1 2 3 \cdots n)$ is $\langle A \rangle$, which is a cyclic group of order $n$. Here, centralizer is the same as stabilizer when the action is conjugation. Hence the orbit, in other words, the number of elements in the conjugacy class of $A$ in $S_n$ is $n!/n=(n-1)!$.