Definition: Let $(V,\Omega)$ be a symplectic vector space, we define $\perp:\Lambda ^k(V^*)\to\Lambda ^{k-2}(V^{\ast})$ by $\perp(\omega)=i_{X_{\Omega}}(\omega)$

here if $(e_1,e_2,...e_n,f_1,f_2,...f_n)$ is basis of symplectic vector space then $\Omega=e_{1}^{*}\wedge f_{1}^{*}+...+e_{n}^{*}\wedge f_{n}^{*} $ and $\ X_\Omega =e_{1}\wedge f_{1}+...+e_{n}\wedge f_{n}$ and $X\in V $ and $\omega_X:=i_X(\omega)$ .

**Question**: Let $Sp(V,\Omega)$ be symplectic group, then I want to see the operator
$q_{\omega}(X)=\perp^2(\omega_X\wedge\omega_X)$ is invariant under $Sp(V,\Omega)$-action
In fact if $S\in Sp(V,\Omega) $

then have we $q_{S^*\omega}(X)=S^{\ast}q_{\omega}(X)$ ?