I want to show the following theorem:
For any Hamilton function $H : M \rightarrow \mathbb{R}$ on some symplectic manifold $M$ and symplectomorphism $f : M \rightarrow M$ we have
$X_{H \circ f} = f^* X_H,$ where $X_g$ is the Hamiltonian vector field with respect to the Hamilton function $g.$
Now, I tried to show this by saying:
$\omega(X_{H \circ f} ,.) = d(H \circ f) = f^*dH= f^* \omega(X_H,.).$
Now I guess at some point, we have to use that $f$ is symplectic, so $f^* \omega = \omega,$ but here we are not dealing with $\omega$ but $\omega(X_H).$
If I'd knew that $f^*\omega(X_H,.) = \omega(f^* X_H,.)$ then this would finish the proof, but I don't know how this can be done.
Well, in order to make thing clear, you can write these equalities explicitly so that you has to see when the assumption : $$f^*\omega= \omega$$ is needed.
Let $p\in M$ and $u\in T_pM$, one has : $$\begin{array}{rcl} \omega_p(X_{H\circ f}(p),u) & = & d_p(H\circ f)(u) \\ & = & d_{f(p)}H(df_p(u)) \\ & = & \omega_{f(p)}(X_H(f(p)),df_p(u)) \\ & = & \omega_{p}(d_{f(p)}f^{-1}(X_H(f(p))),u) \end{array}$$ where the last equality comes from the simple fact that if $f$ is a diffeomorphic symplectomorphism then so is $f^{-1}$.
Hence $$\omega_p(X_{H\circ f}(p),u) = \omega_p(f^*(X_H)(p),u)$$ and you get the result : $$X_{H\circ f} = f^*X_H.$$