Say I define a [time dependent vector field][1] $\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$ as [reversible][2] (also [here][3]) if, for $f(x,y)=(x,-y)$, we have:

$$ f\circ \Psi \circ f =\Psi(-t)=\Psi^{-1}(t).$$

Just to clarify $\Psi:[0,T]\times \mathbb{R}^d\to \mathbb{R}^d$. Now let $\Psi$ be some [Hamiltonian dynamics][4] $\Psi=(q,p)$, where:

$$ \frac{d q}{dt}=\nabla_p H (q,p), \ \ \ ~q(0)=x, $$
$$ \frac{d p}{dt}=-\nabla_q H (q,p), \ \ \ ~p(0)=y, $$
where $H(q,p)=H(q,-p)$. Is it obvious that $\Psi$ is reversible? Can I use https://physics.stackexchange.com/q/528020/ somehow? 

(Note the 2nd equality $\Psi(-t)=\Psi^{-t}(t)$ follows since $(q,p)$ is a flow. https://en.wikipedia.org/wiki/Flow_(mathematics) )

  [1]: https://en.wikipedia.org/wiki/Time_dependent_vector_field
  [2]: https://math.stackexchange.com/questions/1641766/why-is-this-system-reversible-what-does-this-mean
  [3]: https://math.stackexchange.com/questions/2559268/why-is-this-system-reversible-nonlinear-dynamics
  [4]: https://en.wikipedia.org/wiki/Hamiltonian_mechanics