Define $Z_3* Z_3$ to be $\langle a,b\mid a^3,b^3\rangle$. It's easy to see the two paths:
(1) $a \to ab\to aba\to abab \to ababa \to ababab \to \cdots$
and
(2) $a \to aa\to aab\to aaba \to aabaa \to aabaab \to \cdots$
are not equivalent on Cayley graph. But I can't give a strict proof.
My rough idea is to prove there is only one path to connect a vertex on (1), which is after $aa$ and a vertex on (2), which is after $ab$. Otherwise, define such a path as $P$(starting from a vertex $P_1$ after $aa$ on (1) and ending at a vertex $P_2$ after $ab$ on (2): say $P_1\to P_1s_1\to P_1s_1s_2\to \cdots \to P_1 s_1 s_2 \to \cdots \to s_n\to P_2$) , then we can get $P_1s_1s_2\cdots s_n {P_2}^{-1}$ is a loop (relation). However the relations (reduced form) on $\langle a,b \mid a^3, b^3 \rangle$ are generated by $a^3,b^3$, which is a contradiction. So there's only one path. Therefore (1) and (2) are not equivalent.
Can you give me some suggestions on my rough proof, or if you have some other proofs please tell me. Thanks.
The key idea is that the Cayley graph $\Gamma$ is very similar to a tree $T$ known as the "Bass-Serre tree" of the free product $Z_3 * Z_3$. You can get from $\Gamma$ to $T$ in a two step process. First, replace each vertex of $\Gamma$ by two vertices $v,w \in T$ connected by an edge labelled $e$, where the $a$ and $\bar a$ edges of $\Gamma$ are re-attached to $v$ and the $b$ and $\bar b$ edges are re-attached to $w$; let this intermediate graph be denoted $\Gamma_1$. Second, collapse every $a$ edge and every $b$ edge of $\Gamma_1$ to a point, leaving only the $e$ edges. This tree $T$ has one edge orbit and two vertex orbits. Each vertex in the first vertex orbit is stabilized by a conjugate of the order 3 group $\langle a \rangle$, and each vertex in the second vertex orbit is stablized by a conjugate of the order $3$ group $\langle b \rangle$.
The effect of this construction is to produce two quotient maps, each equivariant under the action of $Z_3 * Z_3$, namely $$\Gamma \leftarrow \Gamma_1 \rightarrow T $$ Now take your two rays in $\Gamma$, lift them to $\Gamma_1$, and project them to $T$. You'll obtain two rays in $T$ which go off in different directions almost immediately; and since $T$ is a tree, those rays can never touch again.
And therefore, as you follow each ray out to infinity, its distance to the other ray approaches infinity, so the rays are inequivalent.