= Solution
An element $g$ fixing the first level restricts to an automorphism $g_i$ on each rooted subtree, and composition is coordinatewise. Therefore
$$
\phi(g)=(g_0,g_1,g_2)
$$
is a homomorphism. If all three sections are trivial, $g$ fixes every word, so $\phi$ is injective.
Directly from the recursions,
$$
\phi(b)=(a,1,b),qquad
\phi(aba^{-1})=(b,a,1),qquad
\phi(a^{-1}ba)=(1,b,a).
$$
The generators found in the preceding part therefore have every section in $G$, so $\phi(\operatorname{Stab}_G(1))\subseteq G^3$.
Moreover, every coordinate projection of this image contains both $a$ and $b$, and is therefore onto $G$. Since $a$ acts transitively on the first level, induction shows that $G$ acts transitively on every level of the <rooted tree>. The $n$th level has $3^n$ vertices, so the orders of these finite orbits are unbounded. Hence $G$ is infinite.
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