Solution
= Solution
The preceding cycles show the action on a full <basis>, not only on one direction: $A$ sends $\mathbf e_1\to\mathbf e_2\to\mathbf e_3\to\mathbf e_1$, and $A^{\mathsf T}$ reverses each arrow. Therefore either composition fixes every <basis vector>, and by <linearity> fixes every <vector>. Thus
$$
\boxed{A^{\mathsf T}A=AA^{\mathsf T}=I.}
$$
In particular $A$ is an <orthogonal matrix>, $A^{\mathsf T}=A^{-1}=A^2$, and $A^3=I$.