This is again exercise 1, p. 22 from Lectures on Symplectic Geometry by Ana Cannas da Silva.
Let $(M,\omega)$ be a symplectic manifold and $\alpha \in \Omega^1(M)$ such that $\omega = -d\alpha$. Furthermore let $g: M \to M$ be a symplectomorphism which preserves $\alpha$, i.e. $g^{*}\alpha = \alpha$. Show that $$g_* X = X$$ where $X$ is the unique vector field such that $i_X\omega = -\alpha$.
From my previous post, we get that $X = -\widehat{\omega}^{-1}(\alpha)$, where $\widehat{\omega} : TM \to T^*M$ is defined by $(p,v_p) \mapsto \omega_p(v_p,\cdot)$. Has anyone a hint for me how to show that $g_*X = X$? I mean we have that $$(g_* X)_{g(p)} = dg_p(X_p)$$ But I am not sure how to proceed. I guess I have to use that $g$ is a symplectormorphism and that it preserves $\alpha$, but I am not sure how to use this. Also I am a bit confused by $\omega = -d\alpha$, since until now I did not use this.
Note $g^*\alpha=\alpha$ and $\omega = -d\alpha$ implies $g^* \omega = \omega$. Then for any $Y$,
$$\begin{split} \omega (g_* X,Y) &= (g^* \omega) (X, g^{-1}_* Y) \\ &= \omega(X, g^{-1}_* Y) \\ &= \iota_X \omega (g^{-1}_* Y) \\ &=-\alpha (g^{-1}_* Y) \\ &= -((g^{-1})^* \alpha)(Y) \\ &=-\alpha(Y) \\ &= \iota_X \omega (Y) \\ &= \omega (X, Y). \end{split}$$
Since $Y$ is arbitrary and $\omega$ is nondegenerate, $X = g_*X$.