The automorphism $a$ cyclically permutes the first letter and leaves the remaining suffix unchanged, so $a^3=1$ and $a\ne1$.

In the <section of a rooted-tree automorphism>[section] notation,
$$
b=(a,1,b).
$$
Thus $b^3=(a^3,1,b^3)=(1,1,b^3)$. An automorphism $c$ satisfying $c=(1,1,c)$ fixes every finite word: repeatedly entering the third subtree eventually reaches the end of the word. Hence $b^3=1$. Since $b(\mathbf{00})=\mathbf{01}$, it is nonidentity, and both $a$ and $b$ have order three.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 3 / b / Solution

For an automorphism $g$ of a regular rooted tree and a first-level vertex $i$, the section $g_i$ is the induced automorphism of the rooted subtree at $i$, defined by
$$
g(iv)=g(i)g_i(v).
$$
An automorphism fixing the first level is determined by the tuple of all its sections.


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 Incoming links: Section of a rooted-tree automorphism