In the symmetric group of degree $n$, which elements could possibly commute with the permutation $\sigma$ given by $\sigma(i) = i+1$ if $i < n$; $\sigma(n) = 1$?
Of course, the permutations $e= \sigma^0$, $\sigma$, ..., $\sigma^{n-1}$ all commute with $\sigma$.
Do any other elements too?
On
Your $\sigma$ is actually the cycle $(1 2 \dots n)$, an $n$-cycle. The size of conjugacy class of $\sigma$ is equal to the total number of $n$-cycles, which is equal to $\frac{n!}{n}=(n-1)!$. Thus the number of elements which commute with $\sigma$ is equal to $|S_n|/|Cl_{S_n}(\sigma)|=n!/(n-1)!=n$. You have already found $n$ elements which commute with $\sigma$, and thus they are the only element which commute with $\sigma$. (This is actually an elaboration of the hint given by @YACP).
If $\sigma'=\pi\sigma\pi^{-1}$ where $\pi\in S_n$ then $\sigma'(\pi(i))=\pi(i+1)$ ($i+1$ meant modulo $n$, meaning $n+1\equiv 1$); if $\pi$ and $\sigma$ commute then $\sigma'=\sigma$, i.e. $\pi(i+1)=\pi(i)+1$ (mod $n$), hence $\pi$ is fully determined by $\pi(1)$ and is a power of $\sigma$. So there is nothing else.