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.

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

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.

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

EDIT (too long for a comment) In the above example I really wanted the measure $m$ to be symmetric (i.e., to be preserved by the group inversion). The reason is that you are talking about measures which are stationary simultaneously with respect to $m$ and the reflected measure $\check m$ (the image of $m$ under group inversion). I do not recall any works where measures which are simultaneously $m$- and $\check m$-stationary would be considered (and it might be possible that in some situations all such measures must indeed be invariant - this seems to be an interesting and open question). On the other hand, if $m$ is symmetric, then we are talking about just one stationarity condition, and this situation is quite well understood.

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.

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.

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