Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $X$ be a compact connected $C^\infty$-smooth manifold. Let $F \colon \Omega \times X \to X$ and $\bar{F} \colon \Omega \times X \to X$ be measurable functions such that for each $\omega$, $F(\omega,\,\cdot\,) \colon X \to X$ is a $C^\infty$-diffeomorphism with inverse $\bar{F}(\omega,\,\cdot\,) \colon X \to X$.

Suppose we have a probability measure $\rho$ on $X$, with strictly positive $C^\infty$-smooth density everywhere [as interpreted via charts], such that $$ \rho(A) \ = \ \mathbb{P} \otimes \rho(F^{-1}(A)) \ = \ \mathbb{P} \otimes \rho(\bar{F}^{-1}(A)) $$ for all $A \in \mathcal{B}(X)$. Does it follow that $\mathbb{P}$-almost every $\omega \in \Omega$ has the property that for all $A \in \mathcal{B}(X)$,

$\hspace{50mm} \rho(A) \ = \ \rho(\{x \in X : F(\omega,x) \in A\}) \ $?

**Motivation.** Leaving aside the compactness assumption, take $X=(0,1)$ and $\Omega=\{1,2\}$ with $\mathbb{P}(\{1\})=\mathbb{P}(\{2\})=\frac{1}{2}$. Writing $f=F(1,\cdot)$ and $g=F(2,\cdot)$, I think it should be easy to show that for $\rho$ being the Lebesgue measure,

- $\rho(\cdot)=\mathbb{P} \otimes \rho(\bar{F}^{-1}(\cdot))$ if and only if $g(x)=2x-f(x)$;
- $\rho(\cdot)=\mathbb{P} \otimes \rho(F^{-1}(\cdot))$ if and only if $g$ is the inverse of the map $x \mapsto 2x - f^{-1}(x)$;

and therefore, my very strong intuition (although I don't see an immediate proof) is that we cannot have $\rho(\cdot) = \mathbb{P} \otimes \rho(F^{-1}(\cdot)) = \mathbb{P} \otimes \rho(\bar{F}^{-1}(\cdot))$ unless $f=g=\mathrm{id}_X$: we would require that $x \mapsto 2x-f(x)$ is the inverse of $x \mapsto 2x-f^{-1}(x)$, which is the same as saying that the vertical reflection in $\{y=x\}$ of the diagonal reflection in $\{y=x\}$ of $\mathrm{graph}(f)$ coincides with the diagonal reflection in $\{y=x\}$ of the vertical reflection in $\{y=x\}$ of $\mathrm{graph}(f)$ - and my intuition is that this implies that $f$ is the identity function.

My question above addresses whether, in some sense, this reasoning generalises. I first wondered whether my conjectured result holds in the topological setting, with $X$ being any compact metric space and $\rho$ any probability measure on $X$; but as shown in the answer to this question, the answer is *no*. The counterexample given is very nice, and at the intuitive level (at least under my potentially nonsense intuition) the total disconnectedness of the given counterexample seems important for how my intuitive reasoning about the pair of maps on $(0,1)$ does not extend to this counterexample. So now I am wondering whether having a smooth structure rectifies the situation. [Of course, I might still be wrong even about the case of a pair of maps on $(0,1)$, although this would very much surprise me!]