For $n>1$, and the standard symplectic structure $\omega=\sum dx_i\wedge dy_i$ of $\mathbb{R}^{2n}=\{(x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n)\}$ and for the vector field $X=\partial/\partial_{x_1}$ it is easy to observe that the following vector space is not a Lie algebra, since $Div(\partial/\partial_{x_1})=0$
$$S_{\lambda}(X)=\left\{Y\in \chi^{\infty}(\mathbb{R}^{2n})\mid X.\omega(X,Y)=\lambda Div(X)\omega(X,Y)\right \}$$
But for $n=1$ and $\lambda=1$ it is always a Lie algebra. In fact we have the following obvious fact:
Obvious Fact: Let $(M,\omega)$ be a $2$- dimensional symplectic manifold(i.e: $\omega$ is a volume form on $M$) and $X$ is a vector field on $M$. Then the vector space $$S(X)=\left\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)= Div(X)\omega(X,Y)\right \}$$ is a Lie algebra. Moreover it contains the centralizer $C(X) $
Proof: We apply the well known formula $$d\alpha(X,Y)=X.\alpha(Y)-Y.\alpha(X)-\alpha([X,Y])$$ to $\alpha=i_X(\omega)$. So we conclude that the $S(X)$ in the Obvious Fact is equal to $\{Y\in \chi^{\infty}(M)\mid \omega(X,[X,Y])=0\}$. The later is obviously a Lie algebra containing the centralizer $C(X).$
Remark: For a symplectic manifold $N$ of arbitrary dimension $2n$ it can be shown that the centralizer $C(X)$ of a vector field $X$ is contained in the following vector space:
$$\left\{Y\in \chi^{\infty}(N)\mid X.\omega(X,Y)=(1/n) Div(X)\omega(X,Y)\right \}$$
So in the question of this post one should replace $n$ by $1/n$.
Proof of Remark:
Assume that $[X,Y]=0$. We prove that $X.\omega(X,Y)=(1/n)Div X\omega(X,Y)$. But we need only to prove this formula at all points $p\in N$ with $\omega(X(p),Y(p))\neq 0$. For any such a point $p$, there exist locally a $2$ dimensional symplectic manifold $M$ containing $p$ such that $X,Y$ are tangent to $M$. Now we apply the Obvious Fact above to $M$. We have $X.\omega(X,Y)=Div_{\omega}X.\omega(X,Y) $, where $Div_{\omega} X$ is the divergence of $X$ as a vector field on $M$ with the volume form $\omega$. On the other hand $Div X=(1/n)Div_{\omega} X$ where $Div X$ is the divergence of $X$ as a vector field on the whole manifold $N$ with volume form $\omega^n$.This complete the proof of "Remark".