Are there chaotic maps that commute?

My question is in the title. You can imagine 1D or 2D maps, the simpler the better. Let us say we have chaotic map $$T$$ and chaotic map $$R$$. We need that $$RT(x(n))=TR(x(n))$$.

The simplest example is probably a composition of Bernoulli maps: $$B_a:\quad x \mapsto ax \mod 1,\quad a \in \mathbb{Z}^+,$$

which are going to commute since multiplication is commutative (also under modulo, since $$a$$ is an integer [note 1]).

One could choose, for instance, $$B_2$$ and $$B_5$$ and have $$B_2(B_5(x))=B_5(B_2(x))=B_{10}(x)$$.

An even simpler, but trivial example is an invertible map and its inverse, since by definition they should commute when acting on the same space.

A concrete example is Arnold's cat map $$F$$ and its inverse $$G\equiv F^{-1}$$:

\begin{align} F:& \quad (x,y) \mapsto (2x+y, x+y) \mod 1,\\ G:& \quad (x,y) \mapsto (x-y, 2y-x) \mod 1, \end{align} for which we have that $$FG(x,y)=GF(x,y)=(x,y)$$.

For the commutativity of functions in general, i.e., not necessarily generating chaotic dynamics, Math SE offers some useful answers here and here.

[note 1]
One might expect that the modulo operation could break the commutativity of the (Bernoulli) bit shift maps, but using that $$z \mod 1 = z-N,$$ where $$z=N+y$$, for some integer $$N$$ and real $$y<1$$, we can see that the commutativity holds for $$a,b \in \mathbb{Z}^+$$ and $$x\in[0,1)$$:

\begin{align} B_a(B_b(x)) &=\\ &= [a(bx \mod 1)] \mod 1 = \\ &= [a(bx -N)] \mod 1 = \\ &= [abx -aN] \mod 1 = \\ &= [abx] \mod 1 = \\ &= [abx -bM] \mod 1 = \\ &= [b(ax -M)] \mod 1 = \\ &= [b(ax \mod 1)] \mod 1 = \\ &= B_b(B_a(x)), \end{align}

where we also used that $$z \mod 1 = (z+N) \mod 1$$, for any integer $$N$$, and chose $$M$$ appropriately, i.e., $$M \equiv ax - (ax\mod 1)$$.