Consider the following ODE:

$$\frac{ d \gamma(x,t;\tau)}{d \tau} = R(\gamma(x,t;\tau)) ; \qquad \gamma(x,t,t)=x. $$

$R$ is smooth enough, bounded away from zero and a stationary process. Is there a chance that $\gamma(x,t;\tau)-x$ is stationary with respect to $x$ as well (i.e. $\gamma(x,t;\tau)-x$ has the same distribution as $\gamma(x+y,t;\tau)-x-y$ )

thanks for the any help

PS: one can solve the ODE as follows: let $C(x):=\int_0^x \frac{1}{R(y)} d y$ then $$\gamma(x,t;\tau)=C^{-1}(C(x)-t+\tau),$$

if that helps...