Suppose $S=(s_{ij})$ commutes with it, then

$$\left(\begin{matrix}s_{11}&s_{12}&\cdots&s_{1n}\\s_{21}&s_{22}&\cdots&s_{2n}\\&&\ddots&\\s_{n1}&s_{n2}&&s_{nn}\end{matrix}\right)\left(\begin{matrix}0&1&&\\&0&\ddots&\\&&\ddots&1\\&&&0\end{matrix}\right)=\left(\begin{matrix}0&1&&\\&0&\ddots&\\&&\ddots&1\\&&&0\end{matrix}\right)\left(\begin{matrix}s_{11}&s_{12}&\cdots&s_{1n}\\s_{21}&s_{22}&\cdots&s_{2n}\\&&\ddots&\\s_{n1}&s_{n2}&&s_{nn}\end{matrix}\right)$$

$$\implies\left(\begin{matrix}0&s_{11}&\cdots&s_{1,n-1}\\0&s_{21}&\cdots&s_{2,n-1}\\\vdots&\vdots&&\vdots\\0&s_{n1}&\cdots&s_{n,n-1}\end{matrix}\right)=\left(\begin{matrix}s_{21}&s_{22}&\cdots&s_{2n}\\\vdots&\vdots&&\vdots\\s_{n1}&s_{n2}&\cdots&s_{nn}\\0&0&\cdots&0\end{matrix}\right)$$

Hence $s_{ij}=0$ if $j=1,i\neq1$ or if $i=n,j\neq n$. And $s_{ij}=s_{i+1,j+1}$ for $1\leq i,j\leq n-1$. This means $S$ is of the form

$$\left(\begin{matrix}a_1&a_2&\cdots&a_n\\0&a_1&\ddots&\vdots\\&&\ddots&a_2\\0&&0&a_1\end{matrix}\right)$$

And it is easier to verify all matrices of this form commutes with the Jordan block above.

As a corollary, the set of all matrices having the same form as $S$ above is closed under multiplication.