Can I make a autonomous Hamiltonian diffeomorphism into a non-autonomous one?

Question

Let $H:R^{2n}\to R$ be an autonomous Hamiltonian such that the only constant $1$-periodic orbit is the origin and let $\varphi_H^1$ be the time $1$ map of the flow of it Hamiltonian vector field. Is it possible to find a time dependent Hamiltonian $K_t$ (where the dependence of time ir non-trivial) such that $\varphi_H^1 = \varphi_{K_t}^1$?

Answer 1

See the following exercise [1.4.A from the (great) book "The geometry of the group of symplectic diffeomorphisms" by Leonid Polterovich]:

Reparametrization of flows: Let $f_t, t \in [0, a]$ be a Hamiltonian flow generated by a normalized Hamiltonian $F (x, t)$. Show that $f_{at}, t \in [0, 1]$ is again a Hamiltonian flow generated by $aF (x, at)$. Therefore every Hamiltonian diffeomorphism is in fact a time-one map of some Hamiltonian flow. More generally, show that for every smooth function $b(t)$ with $b(0) = 0$ the flow $f_{b(t)}$ is a Hamiltonian flow whose normalized Hamiltonian equals $\frac{db}{dt}(t)F (x, b(t))$.