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., $M \equiv ax - (ax\mod 1)$.


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