 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. ThusIn particular $A$ is an orthogonal matrix, $A^{\mathsf{T}}=A^{-1}=A^2$, and $A^3=I$.