From time-dependent Hamiltonians to time-dependent symplectic/Poisson structures

Question

Let $(M,\{.,.\})$ be a smooth Poisson manifold, and let $H\in C^\infty(M\times\mathbb{R},\mathbb{R})$.

Question: Does there exist $H_0\in C^\infty(M,\mathbb{R})$ and smooth parameter-dependent Poisson bracket $\{.,.\}_t$ on $M$ such that the following holds? $$ \{H(.,t),f\}=\{H_0,f\}_t,\quad\forall t\in\mathbb{R},\quad\forall f\in C^\infty(M,\mathbb{R}). $$ An essentially equivalent formulation. Let $(M,\omega)$ be a smooth symplectic manifold, and let $H\in C^\infty(M\times\mathbb{R},\mathbb{R})$.

Question: Does there exist $H_0\in C^\infty(M,\mathbb{R})$ and a smooth parameter-dependent symplectic form $\omega_t$ on $M$ such that the following holds? $$ X^\omega_{H(.,t)}=X^{\omega_t}_{H_0}, $$ Here $X$ denotes the Hamiltonian vector field.

Discussion: Basically, I am trying to throw the time dependence from the Hamiltonian to the Poisson structure/symplectic form. It seems to be doable, but I am not completely sure. So instead of reinventing the wheel, I am wondering if someone knows a good reference.

If the answer is negative in this generality, I would appreciate some (hopefully mild) sufficient conditions rather than a pathological counterexample.

A reference would be more helpful than a spelled out proof. Thank you.

Addendum: The answer should be negative in 2 dimensions, since the symplectic form is essentially unique. But in higher dimensions there are plenty of symplectic forms.

Answer 1

Let's denote $H_t$ for the Hamiltonian function.

The support of $X_t^{\omega_t}$ is the support of $dH_t$ independent of the symplectic form $\omega_t$. So if for example the interior of the support of $dH_0$ is empty while the interior of the support of $dH_t$ is not empty, there will be no curve of symplectic forms producing the the same Hamiltonian vector field.

In an analogous way, one should be able to construct a non-symplectic Poisson example.