= Solution
The action on the first level defines a surjective homomorphism
$$
G\longrightarrow\langle(\mathbf0\,\mathbf1\,\mathbf2)\rangle\cong C_3
$$
that sends $a$ to the displayed cycle and $b$ to the identity. Its kernel is therefore the normal closure of $b$, proving that $\{b\}$ normally generates $\operatorname{Stab}_G(1)$.
Apply the <Reidemeister–Schreier theorem> with transversal $\{1,a,a^{-1}\}$. The generators arising from $a$ are trivial, while those arising from $b$ are
$$
b,qquad aba^{-1},qquad a^{-1}ba.
$$
Consequently these three elements generate $\operatorname{Stab}_G(1)$.
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