Let $G$ be a finite group with $n = |G|$ elements. By Cayley's theorem for finite groups, we have an injective homomorphism of groups: $$ \pi : G \rightarrow S_n, \quad g \mapsto \pi(g) $$ where each group element $g$ is mapped to the permutation of $G$ by left multiplication: $$ \pi(g): G \rightarrow G, \quad x \mapsto g \cdot x. $$
Let $T_0:=$ Cayley table of $G$ and let $C_{i+1} := G \cdot T_{i}$, where
$$G \cdot T_{i}$$
is to be understood as $G$ acting on the cells of the iterated Cayley table of depth $i$, which for $G = C_2$ is the same as the Thue-Morse plane.
It is easier to explain this with examples:
$G = C_2$ and $i=0,1,2$
$$\left(\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right) \left(\begin{array}{rrrr} 1 & 2 & 2 & 1 \\ 2 & 1 & 1 & 2 \\ 2 & 1 & 1 & 2 \\ 1 & 2 & 2 & 1 \end{array}\right) \left(\begin{array}{rrrrrrrr} 1 & 2 & 2 & 1 & 2 & 1 & 1 & 2 \\ 2 & 1 & 1 & 2 & 1 & 2 & 2 & 1 \\ 2 & 1 & 1 & 2 & 1 & 2 & 2 & 1 \\ 1 & 2 & 2 & 1 & 2 & 1 & 1 & 2 \\ 2 & 1 & 1 & 2 & 1 & 2 & 2 & 1 \\ 1 & 2 & 2 & 1 & 2 & 1 & 1 & 2 \\ 1 & 2 & 2 & 1 & 2 & 1 & 1 & 2 \\ 2 & 1 & 1 & 2 & 1 & 2 & 2 & 1 \end{array}\right)$$
$G=C_3$ and $i=0,1$
$$ \left(\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right) \left(\begin{array}{rrrrrrrrr} 1 & 2 & 3 & 2 & 3 & 1 & 3 & 1 & 2 \\ 2 & 3 & 1 & 3 & 1 & 2 & 1 & 2 & 3 \\ 3 & 1 & 2 & 1 & 2 & 3 & 2 & 3 & 1 \\ 2 & 3 & 1 & 3 & 1 & 2 & 1 & 2 & 3 \\ 3 & 1 & 2 & 1 & 2 & 3 & 2 & 3 & 1 \\ 1 & 2 & 3 & 2 & 3 & 1 & 3 & 1 & 2 \\ 3 & 1 & 2 & 1 & 2 & 3 & 2 & 3 & 1 \\ 1 & 2 & 3 & 2 & 3 & 1 & 3 & 1 & 2 \\ 2 & 3 & 1 & 3 & 1 & 2 & 1 & 2 & 3 \end{array}\right) $$
$G=C_2 \times C_2$ and $i=0,1$
$$ \left(\begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \end{array}\right) \left(\begin{array}{rrrrrrrrrrrrrrrr} 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 \\ 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 \\ 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 \\ 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 \\ 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 \\ 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 & 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 & 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 \\ 2 & 1 & 4 & 3 & 1 & 2 & 3 & 4 & 4 & 3 & 2 & 1 & 3 & 4 & 1 & 2 \\ 1 & 2 & 3 & 4 & 2 & 1 & 4 & 3 & 3 & 4 & 1 & 2 & 4 & 3 & 2 & 1 \end{array}\right) $$
We can use this process to create Truchet tilings as follows:
Let $T_i$ be a Cayley table of depth $i$ of some finite group.
The entry $(k,l)$ of $T_i$ is a natural number $t_{k,l}$ which for a Truchet tiling with cells being in $s$ states $0,1,\cdots,s-1$ (for example $s=2,4,5,6$), we can view the number $t_{k,l}$ modulo $s$ :
$$t_{k,l} \mod (s)$$
and so we can construct a Truchet tiling from the iterated Cayley table.
The question which occupied me, is:
Q: For which finite group $G$ and for which $s \in \{2,4,5,6\}$, do we get a "non-periodic" tiling with Truchet cells, when we iterate the depth of the Cayley table $i$ to infinity?
(Feel free to interpret "non-periodic" in a sense, which comes close to "non-periodic"...)
Here are some images:
A finite group with order $12$ and $s=2$:
$C_3$ Cayley table ($s=4$)
Klein Four group (iterated table, $s=5$):
$S_3$ group (iterated table, $s=5$):
