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".