A linear operator commuting with all such operators is a scalar multiple of the identity.

For a basis-free answer, consider $S \in L(V)$ given by $S x = f(x) v$ for some vector $v$ and some linear functional $f$ on V. Then $T S x = f(x) T v = S T x = f(T x) v$ for any x. In particular, as long as a nontrivial linear functional $f$ on $V$ exists, there is $x$ such that $f(x) \ne 0$, and then $T v = \alpha v$ for all $v$, where $\alpha = f(T x)/f(x)$. This works even for infinite-dimensional spaces, although I think in general you need the Axiom of Choice to get a nontrivial linear functional on a vector space.

Suppose $TS = ST$ for every $S$. Show that $Tv = a_{v}v$ for every $v\in V$ where $a_v$ could depend on $v$. In other words, show that $v$ and $Tv$ are linearly dependent for each $v \in V$.

Suppose for contradiction that they are linearly independent. Since $(v, Tv)$ is linearly independent, it can be extended to a basis $(v,Tv, u_1, \dots, u_n)$ of $V$. So define $S$ as following: $Sv = v$, $S(Tv) = v$ and $S(u_1) = 0, \dots, S(u_n) = 0$. Then, $Tv = TSv = STv = v$. Hence $v$ and $Tv$ are linearly dependent, which is a contradiction. Then you have to show uniqueness.

In general, when one has a condition of the form "$A$ is a blah if and only if for every $B$ this happens", the "if" direction can often be established by selecting suitably/cleverly chosen $B$ that show everything works.

This is just such a situation.

Let $\beta = \{\mathbf{v}_i\}_{i\in I}$ be a basis for $\mathbf{V}$. For each $i,j\in I$, let $S_{ij}$ be the linear operator on $\mathbf{V}$ given by $$S_{ij}(\mathbf{v}_k) = \left\{\begin{array}{ll} \mathbf{v}_j &\mbox{if $k=i$,}\\ \mathbf{v}_i &\mbox{if $k=j$,}\\ \mathbf{0} &\mbox{if $k\neq i$ and $k\neq j$.} \end{array}\right.$$ That is: for $i\neq j$, $S_{ij}$ exchanges $\mathbf{v}_i$ and $\mathbf{v}_j$, and maps all other basis elements to $\mathbf{0}$. And $S_{ii}$ maps $\mathbf{v}_i$ to itself, and all other basis elements to $\mathbf{0}$. These are our "suitably chosen" $S$.

Now consider $S_{ii}T(\mathbf{v}_j)$ and $TS_{ii}(\mathbf{v}_j)$ first to get information about what $T$ does to $\beta$; then consider $S_{ij}T(\mathbf{v}_j)$ and $TS_{ij}(\mathbf{v}_j)$ for $i\neq j$.