Let $P, P'$ be symplectic(resp. Hamiltonian) fibrations over a base $B$.
If two $P, P'$ are continuously symplectic fibration(resp. Hamiltonian fibration) isomorphic, then are they also smoothly symplectic fibration(resp. Hamiltonian fibration) isomorphic?
In here, a symplectic fibration $E$ over a base $B$ is a smooth bundle whose fiber is a symplectic manifold $(F, \omega)$, and the structure group is $\mathrm{Symp}(F, \omega)$.
A Hamiltonian fibration $E$ over a base $B$ is a symplectic fibration that admits global coupling $2$-form $\Omega$ on $E$ such that $\Omega |_{\ker \mathrm{d}_p \pi} = \omega_p$ at each fiber $\pi^{-1}(p)$, and there is a symplectic connection such that the connection induced by $\Omega$ is equal to the connection. (or equivalently, there is a symplectic connection such that the holonomy along any contractible loop is a Hamiltonian group.)
Two fibrations are symplectic fibration isomorphic, if there is a bundle map $\phi : P \to P'$ such that $\phi^*_p \omega_p = \omega_p$ at each fiber $\pi^{-1}(p)$.
Two fibrations are Hamiltonian fibration isomorphic, if they are symplectic fibration isomorphic, and the fibration isomorphism preserves the coupling $2$-form.
I guess there is a proof by the theory of classifying space, but I don't know where is a good starting point. I would like to know which references are good to cite and prove this result.
