Hamiltonian diffeomorphisms generated by periodic Hamiltonian function Let $\phi_{t}$ be a family of Hamiltonian diffeomorphisms on a symplectic manifold $(M,\omega)$, generated by a family of Hamiltonians $H(x,t)$. I'm trying to show that the family of Hamiltonians is periodic with period $1$ with respect to $t$ (in other words $H(x,t) = H(x,t+1)$ for all $x$) if and only if the family of diffeomorphisms satisfies $\phi_{t+1} = \phi_{t} \circ \phi_{1}$.
I know from the definition that $\frac{d}{dt}\phi_{t}(x) = X_{H}(\phi_{t}(x))$, where $X_{H}$ is the Hamiltonian vector field and that $i_{X_{H}}\omega = dH$. The symplectic form is compatible with this vector field in the sense that $\omega(X_{H}(x), \cdot) = dH$.
I tried to differentiate the equation $\phi_{t+1}=\phi_{t} \circ \phi_{1}$ and I obtain the following:
$$X_{H}(\phi_{t+1}(x)) = X_{H}(\phi_{t}(\phi_{1}(x))) \cdot X_{H}(\phi_{1}(x))$$
But I don't see how to proceed or finish the proof for either of the implications. How can I prove this?

For one implication, assume that $H(x,t)=H(x,t+1)$. To prove that $\phi_{t+1}=\phi_{t}\circ\phi_{1}$, you can show that both sides solve the initial value problem
$$
\begin{cases}
\frac{d}{dt}\psi_{t}=X_{H_t}\circ\psi_{t}\\
\psi_{0}=\phi_1
\end{cases}.
$$
Indeed, on one hand we have
$$
\frac{d}{dt}\phi_{t+1}=X_{H_{t+1}}\circ\phi_{t+1}=X_{H_t}\circ\phi_{t+1},
$$
using that $H_{t+1}=H_t$. On the other hand we have
$$
\frac{d}{dt}\phi_{t}\circ\phi_{1}=X_{H_t}\circ\phi_t\circ\phi_1.
$$
The other implication doesn't seem true to me. Information about the flow tells you something about the Hamiltonian vector field, which does not determine the Hamiltonian function uniquely. For instance, take $$(\mathbb{R}^{2},dx\wedge dy)\ \ \text{and}\ \ H(x,y,t)=y+t.$$ Then
$
X_{H}=\partial_{x}
$
is time-independent, so its flow satisfies $\phi_{t+1}=\phi_t\circ\phi_1$. But $H(x,y,t+1)\neq H(x,y,t)$.
