Group Theory – Finding the Centralizer of a Permutation

group-theorypermutations

I need to find the centralizer of the permutation $\sigma=(1 2 3 … n)\in S_n$.
I know that:
$C_{S_n}(\sigma)=\left\{\tau \in S_n|\text{ } \tau\sigma\tau^{-1}=\sigma\right\}$
In other words, that the centralizer is the set of all the elements that commute with $\sigma$, and I also know that if two permutations have disjoint cycles it implies that they commute, but the thing is; there are no $\tau\in S_n$ s.t. $\tau$ and $\sigma$ have disjoint cycles, since $\sigma=(1 2 3…n)$.
So can I conclude that $\sigma$ does not commute with any other $\tau$ in $S_n$ (besides $id$ of course)?
I guess my question reduces to: is the second direction of the implication mentioned above also true? meaning, if two permutation commute, does it imply that they have disjoint cycles?
If the answer is no, how else can I find the $C_{S_n}(\sigma)$?

By the way, on related subject, I noticed that if an element $g$ of a group $G$ is alone it its conjugacy class, it commutes with all elements in $G$.
What does it mean, intuitively, for an element to share its conjugacy class with another element? does it mean it "almost" commute with everyone in the group?
Is it true that the bigger the conjugacy class, the lesser its members commutes with others in the group?

Best Answer

Hint: conjugacy in $S_n$ leaves cycle types in tact: $\tau^{-1}(1 2 3 \dots n)\tau=(\tau(1) \tau(2)\tau(3) \cdots \tau(n))$.