Let's say we have two densities $p_t$ and $q_t$ on $\mathbb{R}^d$ which have the same time-evolution in the Liouville (Fokker-Planck with no noise terms) sense. That is,
$ \frac{ \partial p_{t}(x) }{\partial t } = - \nabla \cdot \left( f_t(x) p_{t}(x) \right) \notag \qquad \text{and} \qquad \frac{ \partial q_{t}(x) }{\partial t } = - \nabla \cdot \left( g_t(x) q_{t}(x) \right) \notag$
where $\nabla \cdot$ corresponds to the divergence operator. It is easy to see that although $f_t \neq g_t$ we can have the same evolution which means that the solutions $p_t(x)$ and $q_t(x)$ are the same if the initial conditions $p_0$ (say standard Gaussian) are the same. So far so good.
Note that for $p_t$ and $q_t$, there exists corresponding stochastic processes $X_t$ and $Y_t$, respectively, which follow the deterministic ODEs
$ \frac{d X }{d t } = f_t(X), \quad X(0) \sim p_0(x) \notag \qquad \text{and} \qquad \frac{d Y }{d t } = g_t(Y), \quad Y(0) \sim p_0(x) \notag$
see "Interacting particle solutions of Fokker–Planck equations through gradient–log–density estimation".
Question: Although $f_t \neq g_t$, does the fact the the true density follow the same PDE imply that the two stochastic processes $X_t$ and $Y_t$ are somewhat equivalent? I don't see any sensible reason why that should not be the case but I guess I am assuming some regularities on $f_t$ and $g_t$ to ensure uniqueness of the PDE solutions?