Let $(M,\omega)$ be a $2n$ dimensional symplectic manifold and $X$ is a smooth vector field on $M$. Consider the following subvector space of $\chi^{\infty}(M)$: $$S(X)=\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)=nDiv(X)\omega(X,Y)\}$$.

This vector space contains the Lie algebra $C(X)=\{Y\in \chi^{\infty}(M)\mid [X,Y]=0\}$. It also contains the Lie algebra $M(X)=\{fX\mid f\in C^{\infty}(M)\}$

Is $S(X)$ a Li subalgebra of $\chi^{\infty}(M)$?

If the answer is no, is the Lie algebra generated by $S(X)$ equal to the lie algebra generated by $C(X)$ and $M(X)$?

Motivated by the fact that "triviality of centralizer of a vector field is somehow a generic situation" is it true to say that for a generic vector field $X$ we have $S(X)=M(X)$?