Let $(X_t)_{t\ge 0}$ and $(Y_t)_{t\ge 0}$ be a pair of stochastic processes taking values in $\mathbb{R}^n$ and in $\mathbb{R}^m$; defined on a filtered probability spaces $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P})$. Suppose that $$ \begin{aligned} X_t = & x + \int_0^t \, \mu(s,X_s)ds + \int_0^t\, \sigma(s,X_s)dW_s\\ Y_t = & y + \int_0^t \alpha(X_s,Y_s)ds + \int_0^t \, \beta(s,X_s,Y_s)dB_s \end{aligned} $$ where $(W_t)_{t\ge 0}$ and $(B_t)_{t\ge 0}$ are independant $\mathcal{F}$-adapted Brownian motions, and $\mu,\sigma,\alpha,\beta$ are uniformly (in time) Lipschitz smooth functions; such that $(X_t,Y_t)_{t\ge 0}$ has a unique strong solution.

Let $(\mathcal{F}^Y_t)_{t\ge 0}$ denote the right-continuous completed filtration generated by $(Y_t)_{t\ge 0}$ and let $\mathbb{P}(X_t|\mathcal{F}_t^Y)$ denote the regular conditional distribution of $X_t$ given $\mathcal{F}_t^Y$. By the Markovianity of both processes, we have that
$$ \mathbb{P}(X_t|\mathcal{F}_t^Y) = f(t,Y_t), $$ for some Borel function $f:[0,\infty)\times \mathbb{R}^m\rightarrow \mathcal{P}(\mathbb{R}^n)$. Under some integrability assumptions, we can even deduce that $f$'s image is in $\mathcal{P}_1(\mathbb{R}^n)$ equipped with the 1-Wasserstein metric $W_1$.

Under what conditions on the dynamics $\mu,\sigma,\alpha,\beta$ can we deduce that, for every $T>0$ the map $f|_{[0,T]\times \mathbb{R}^m}$ is Lipschitz?

Let $(X_t)_{t\ge 0}$ and $(Y_t)_{t\ge 0}$ be a pair of stochastic processes taking values in $\mathbb{R}^n$ and in $\mathbb{R}^m$; defined on a filtered probability spaces $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P})$. Suppose that $$ \begin{aligned} X_t = & x + \int_0^t \, \mu(s,X_s)ds + \int_0^t\, \sigma(s,X_s)dW_s\\ Y_t = & y + \int_0^t \alpha(X_s,Y_s)ds + \int_0^t \, \beta(s,X_s,Y_s)dB_s \end{aligned} $$ where $(W_t)_{t\ge 0}$ and $(B_t)_{t\ge 0}$ are independant $\mathcal{F}$-adapted Brownian motions, and $\mu,\sigma,\alpha,\beta$ are uniformly (in time) Lipschitz smooth functions; such that $(X_t,Y_t)_{t\ge 0}$ has a unique strong solution.

Let $(\mathcal{F}^Y_t)_{t\ge 0}$ denote the right-continuous completed filtration generated by $(Y_t)_{t\ge 0}$ and let $\mathbb{P}(X_t|Y_t)$ denote the regular conditional distribution of $X_t$ given the $\sigma$-algebra generated by $Y_t$ (As Llya pointed out, this may be different from the $\sigma$-algebra generated by the process $(Y_s)_{s\ge 0}$ up to time $t$).
By the Markovianity of both processes, we have that
$$ \mathbb{P}(X_t|Y_t) = f(t,Y_t), $$ for some Borel function $f:[0,\infty)\times \mathbb{R}^m\rightarrow \mathcal{P}(\mathbb{R}^n)$. Under some integrability assumptions, we can even deduce that $f$'s image is in $\mathcal{P}_1(\mathbb{R}^n)$ equipped with the 1-Wasserstein metric $W_1$.

Under what conditions on the dynamics $\mu,\sigma,\alpha,\beta$ can we deduce that, for every $T>0$ the map $f|_{[0,T]\times \mathbb{R}^m}$ is Lipschitz?