These are just an attempt to reformulate everything in a more convenient way.

1. Let us define the following function on nonnegative integers (the definition can be simplified a bit, but I prefer to present it in this form): $$ f(n)=\begin{cases} 4n, & x=6n\; \text{ or }\;x=6n+1;\cr 4n+2, & x=6n+2\; \text{ or }\;x=6n+3;\cr 2n, & x=6n+4\; \text{ or }\;x=6n+5. \end{cases} $$ Then $f(x) < x$ for all $x>2$ and $x=1$, and $f(x)$ lies in the orbit of $x$. Moreover, it can be checked straightforwardly that for every two numbers interchanged by one of the generators, they have a common image under some iterations of $f$. This means that one orbit consists of all the numbers coming to $0$ after a sufficient number of iterations, and the other orbit consists of those coming to $2$.

2. Now let us check what happens in this process. Surely one may concentrate only on the even numbers, since $f(2n)=f(2n+1)$ is always even. Thus we change the variable by $y=x/2$ and introduce the function $$ g(y)=\frac{f(2y)}2=\begin{cases} 2n, & y=3n;\cr 2n+1, & y=3n+1;\cr n, & y=3n+2\cr \end{cases} = \begin{cases} \lfloor y/3\rfloor, & 3\;\big|\; (y+1);\cr \lceil 2y/3\rceil, & \text{otherwise.} \end{cases} $$ Considering the preimages under $g$, we see that each $y$ generates the numbers $3y+2$ and $\lfloor 3y/2\rfloor$ lying in the $y$'s orbit, and all the numbers are generated by such process from $0$ and $1$ (or from $1$ and $2$), thus partitioning into two orbits.

3. Now, if we denote by $\mu(n)$ the density of, say, elements of the first orbit on $[2n,3n)$, then we obtain $$ \mu(3n)=\frac{\mu(n)+2\mu(2n)}3. $$ It seems that this, together with the observation that $\mu(n)$ and $\mu(n+1)$ are close to each other, is enough to see that the limiting density exists.

These are just an attempt to reformulate everything in a more convenient way.

1. Let us define the following function on nonnegative integers (the definition can be simplified a bit, but I prefer to present it in this form): $$ f(n)=\begin{cases} 4n, & x=6n\; \text{ or }\;x=6n+1;\cr 4n+2, & x=6n+2\; \text{ or }\;x=6n+3;\cr 2n, & x=6n+4\; \text{ or }\;x=6n+5. \end{cases} $$ Then $f(x) < x$ for all $x>2$ and $x=1$, and $f(x)$ lies in the orbit of $x$. Moreover, it can be checked straightforwardly that for every two numbers interchanged by one of the generators, they have a common image under some iterations of $f$. This means that one orbit consists of all the numbers coming to $0$ after a sufficient number of iterations, and the other orbit consists of those coming to $2$.

2. Now let us check what happens in this process. Surely one may concentrate only on the even numbers, since $f(2n)=f(2n+1)$ is always even. Thus we change the variable by $y=x/2$ and introduce the function $$ g(y)=\frac{f(2y)}2=\begin{cases} 2n, & y=3n;\cr 2n+1, & y=3n+1;\cr n, & y=3n+2\cr \end{cases} = \begin{cases} \lfloor y/3\rfloor, & 3\;\big|\; (y+1);\cr \lceil 2y/3\rceil, & \text{otherwise.} \end{cases} $$ Considering the preimages under $g$, we see that each $y$ generates the numbers $3y+2$ and $\lfloor 3y/2\rfloor$ lying in the $y$'s orbit, and all the numbers are generated by such process from $0$ and $1$ (or from $1$ and $2$), thus partitioning into two orbits.

3. Now, if we denote by $\mu(n)$ the density of, say, elements of the first orbit on $[2n,3n)$, then we obtain $$ \mu(3n)=\frac{\mu(n)+2\mu(2n)}3. $$ It seems that this, together with the observation that $\mu(n)$ and $\mu(n+1)$ are close to each other, should suffice to see that the limiting density exists.