# Proving Hamiltonian symplectomorphism commutes with given symplectomorphism

I'm trying to prove this identity which is mentioned at the very beginning of this paper by Dostoglou and Salamon "Self-Dual instants and holomorphic curves".

Let $$(M,\omega)$$ be a closed symplectic manifold and let $$\phi$$ a symplectomorphims $$\phi \colon M \to M$$. Let $$H \colon \Bbb R\times M\to M$$ be a smooth time-dependant Hamiltonian function such that $$H_s=H_{s+1}\circ \phi$$. The Hamiltonian symplectomorphisms $$\psi_s \colon M \to M$$ generated by $$H$$ are defined by $$\frac{d}{ds}\psi_s =X_s\circ \psi_s, \ \ \psi_0=Id, \ \ \omega(X_s, \cdot ) = dH_s(\cdot)$$ They satisfy $$\psi_{s+1}\circ \phi_H=\phi \circ \psi_s$$ where $$\phi_H := \psi_1^{-1}\circ \phi$$.

Equivalently we must show $$\psi_s\circ \phi =\phi \circ \psi_s$$ It's clear that the equation is satisfied at time $$s=0$$, and if we compute the derivative w.r.t. $$s$$ at $$s=0$$ we get on both sides $$X_1\circ \phi$$, where I used the fact that $$H_s=H_{s+1}\circ \phi$$. But from here I don't know what else can I say.

Can someone help me figuring out this equality?

Rewriting, you need to prove that $$\psi_{s+1}\circ \psi_1^{-1} = \phi\circ\psi_s\circ\phi^{-1}.$$ This is not actually equivalent to what you wrote - I think you are assuming that $$\psi_s$$ is a one parameter group, and so $$\psi_{s+1}\circ\psi_1^{-1} = \psi_s$$, but this is not true for the flow of a time-dependent vector field.
Regardless, let $$f_s = \psi_{s+1}\circ\psi_1^{-1}$$, and $$g_s = \phi\circ\psi_s\circ \phi^{-1}$$. We want to show these two diffeomorphisms are the same. Firstly, both equal the identity at $$s=0$$. Now for any $$p\in M$$ $$\frac{d}{ds}f_s(p) = \frac{d}{ds}\psi_{s+1}(\psi_1^{-1}(p)) = X_{s+1}(\psi_{s+1}(\psi_1^{-1}(p))) = X_{s+1}(f_s(p)).$$ Also, starting from $$H_s = H_{s+1}\circ \phi$$, it's not difficult to show that for any $$p\in M$$, $$T_p\phi\,(X_s(p)) = X_{s+1}(\phi(p)).$$ It follows that $$\frac{d}{ds}g_s(p) = \frac{d}{ds}(\phi\circ\psi_s\circ\phi^{-1})(p) = T_{\psi_s(\phi^{-1}(p))}\phi\, \left(X_s(\psi_s(\phi^{-1}(p)) \right) \\= X_{s+1}(\phi(\psi_s(\phi^{-1}(p))) = X_{s+1}(g_s(p)).$$ So both $$f_s$$ and $$g_s$$ satisfy the same 1st-order ODE with the same initial condition. Therefore they are identical.