Discontinuity is not crucial at all in the example given in the answer to [this question][1], and the same phenomenon is present in the smooth setup as well. Namely, the free group can be replaced with $SL(2,\mathbb R)$ (or any non-compact semi-simple Lie group) and the space of infinite words with the boundary circle of the hyperbolic plane (or the "flag space" of the associated Riemannian symmetric space).

More precisely, let $G$ be the group $SL(2,\mathbb R)$. It acts in the standard way by linear fractional transformations on the augmented complex plane and preserves the augmented real line $S^1$ ($\equiv$ the boundary circle of the hyperbolic plane in the upper half-plane model). Let $K=SO(2)\subset G$ be the group of rotations, and let $m$ be a bi-$K$-invariant probability measure on $G$ (in particular, $m$ is preserved by the group inversion). Now one can put $(\Omega,\mathbb P)=(G,m)$ and $X=S^1$ with the map $\Omega\times X\to X$ being the aformentioned boundary action. The unique $K$-invariant measure $\rho$ on $S^1$ (the "Haar measure") is then $m$-stationary, but not $G$-invariant.

PS Stationarity alone is not sufficient to imply invariance (as these examples illustrate). It is reversibility that is essentially equivalent to invariance.


  [1]: https://mathoverflow.net/questions/343977/if-a-probability-measure-is-stationary-in-both-forward-time-and-reverse-time-do