Let $X^*$ be the free monoid on a set $X$. Now define the graph with vertices $X^*$, and the edges $E \subset (X^*)^2$ recursively as follows:
- $(1, 1) \in E$.
- If $(w, w') \in E$, then $(w', w' \cdot x) \in E$ for all $x \in X$.
- If $t: [n] \to X^*$ is a path in $E$, meaning $s \in [n-1]$ implies $(t(s), t(s+1)) \in E$, we must have that $(t(n), t(n) \cdot t(i)) \in E$ for $i \in [n]$.
Given $x \in X^*$, what is the length of the shortest path from $1$ to $x$?
Note that this is not a homomorphism (call it $|\cdot| : X^* \to \mathbb N)$ since $|x^3| = |x^4| = 3$ with $X^*$ being the free monoid on the terminal set $\{x\}$.
My understanding. It looks like you are mostly defining the Cayley graph of the free monoid $X^*$ of base $X$. The vertices of this graph are the elements of $X^*$ and the edges are of the form $(u, ux)$, where $u \in X^*$ and $x \in X$. Your definition also includes the edge $(1,1)$, but it does not change much.
In particular, if $X = \{x\}$, then $$ E = \{(1,1)\} \cup \{(x^n, x^{n+1}) \mid n \geqslant 0\} $$ so the shortest part from $1$ to $x^n$ has length $n$. It is the length of the path $$ 1 \xrightarrow{x} x \xrightarrow{x} x^2 \quad \dotsm \quad x^{n-1} \xrightarrow{x} x^n $$ in the Cayley graph.
Coming back to your comments, $\Gamma^3(1)$ is the set of neighbours of $\Gamma^3(1)$ in the Cayley graph. Since $\Gamma^2(1) = \{1, x, x^2\}$, $\Gamma^3(1) = \{1, x, x^2, x^3\}$. In particular $x^4$ is not a neighbour of any element of $\Gamma^2(1)$.