Let $\mu$ is stationary distribution of discrete random dynamical system define as $$\mu=\begin{cases} 1 & p_1, \\ 2 & p_2,\\ 3 & p_3, \\ \end{cases}$$ where $p_i\in \left(0,1\right)$, for $i=1,2,3,$ and $p_1+p_2+p_3=1$.
The question is, is there discrete random dynamical system
$$x_{n+1}= \begin{cases} f(x_n) & q, \\ g(x_n) & 1-q, \end{cases}$$ where $q\in(0,1)$ and function $f$, $g$ $\{1,2,3\}\rightarrow\{1,2,3\}$ with this stationary distribution $\mu$?
My opinion is that it is impossible to make discrete random dynamical system to use only two function $f$ and $g$ with this stationary distribution $\mu$. It is clear to me that it is possible with three function $f$, $g$ and $h$. But it is not my task. I have been thinking like this, stationary distribution of random dynamical system is distribution which hold this $$\mu(A)=q\cdot\mu(f^{-1}(A))+(1-q)\cdot\mu(g^{-1}(A)),$$ for any $A\subseteq\{1,2,3\}$. There are finite number of function from $\{1,2,3\}$ to $\{1,2,3\}$, so there are $3^3=27$ different possibilities how to choose function $f$ and the same for function $g$. So there are $27\cdot27=729$ possibilities how to choose these two functions. Of course there are some pairs which are obviously not good, for example $f(x)=1$ and $g(x)=1$ $\forall x\in\{1,2,3\}$ and etc. But it is still a lot of work to check all possibilities for functions $f$ and $g$ and different subsets of set $\{1,2,3\}$. There are some elegant "proof/solution" of this problem. I have been thinking about proof by contradiction but I have no idea. Any help will be appreciated, thank you very much.