Show that if $\Phi:M\to M$ is a diffeomorphism, then the induced map $\Phi^*:T^*M\to T^*M$ is a symplectomorphism.
Presumably, $T^*M$ is equipped with the canonical 2-form $\omega$. To show $\Phi^*$ is a symplectomorphism, we must show it is a diffeomorphism and $(\Phi^*)^*(\omega)=\omega$. I am having trouble with the second part. The map $$(\Phi^*)^*:T^*(T^*M)\to T^*(T^*M)$$ takes a 1-form $\theta$ to $(\Phi^*)^*(\theta)$, where $$(\Phi^*)^*(\theta)X=\theta((\Phi^*)_*X)=\theta(X\circ\Phi^*)$$ for $X\in T(T^*M)$. And this map extends to two forms in the usual way: $$(\Phi^*)^*(\omega)(X,Y)=\omega((\Phi^*)_*X,(\Phi^*)_*Y)=\omega(X\circ\Phi^*,Y\circ\Phi^*)$$ for $X,Y\in T(T^*M)$. So at first glance, it seems like I need to show that $\Phi^*$ is the identity, but surely this is not the case in general. So where have I gone wrong?
I believe I have identified a flaw in my understanding of the pushforward that has allowed me to solve this problem. Perhaps someone can verify that this is correct.
For $X\in T(T^*M)$ and $(x,\phi)\in T^*M$, we have that $$(\Phi^*)^*(\theta)(X)_{(x,\phi)}=\theta((\Phi^*)_*X)_{(x,\phi)}=\phi(\pi_*((\Phi^*)_*X)).$$ But for $f:M\to\mathbb{R}$, $$\pi_*((\Phi^*)_*(X))f=(\Phi^*)_*(X)(f\circ\pi)=X(f\circ\pi\circ\Phi^*)$$ $$=X(f\circ\Phi\circ\pi)=\pi_*(X)(f\circ\Phi)=\Phi_*(\pi_*(X))f.$$ i.e., $$\pi_*((\Phi^*)_*(X))=\Phi_*(\pi_*(X)).$$ Then $$(\Phi^*)^*(\theta)(X)_{(x,\phi)}=\phi(\pi_*((\Phi^*)_*X))=\phi(\Phi_*(\pi_*X))$$ $$=\Phi^*(\phi)(\pi_*X)$$ and I believe that this is the 1-form $\theta$ but on the other copy of $T^*M$.