Discontinuity is not crucial at all in the example given in the answer to this question, 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.