Symplectomorphism Preserves Cotangent fibrations Let $M$ be a manifold with local coordinates $x^1,\dots,x^n$ and $T^\ast M$ the cotangent bundle with induced coordinates $x^1,\dots,x^n,\xi_1,\dots,\xi_n$ . Let $\alpha$ be the tautological one form on $T^\ast M$. That is, $\alpha\in \Gamma(T(T^\ast M))$ where $\alpha=\xi_idx^i$. Let $\omega=-d\alpha=dx^i\wedge d\xi_i$. We have that $(T^\ast M,\omega)$ is a symplectic manifold. Let $g:T^\ast M\to T^\ast M$ be a symplectomorphism that preserves $\alpha$ (i.e. $g^\ast\alpha=\alpha$). 
I'm trying to show that if for some $(p,\sigma_p)\in T^\ast_p M$ we have $g(p,\sigma_p)=(q,\eta_q)$, then $g(T^\ast_p M)=T^\ast_q M$. 
So far this is what I have:
Let $V\in\Gamma(T(T^\ast M))$ be the symplectic dual of $\alpha$. That is, $\omega(V,\cdot)=\alpha$. Let $\theta$ denote the flow of $V$. I can show that $g\circ\theta_t=\theta_t\circ g$ for all $t\in\mathbb{R}$ or equivalently, that $g_\ast V=V$. I can show that each $\theta_t$ preserves $T_p^\ast M$ for arbitrary $p\in M$. That is, I can show that $\theta_t(T^\ast_p M)=T^\ast_p M$. This follows from actually computing the flow, which turns out to be $\theta_t(p,\mu_p)=(p,e^t\mu_p)$. How do I use these facts to show what I wrote above? Am I missing something obvious? Any help would be greatly appreciated.

Just to take this off the unanswered list, the answer to this question was provided here 
https://mathoverflow.net/questions/175807/lifting-a-diffeomorphism-to-the-cotangent-bundle
