How many $2\times2$ orthogonal matrices $A$ with $A^3=I$ and $A^2=A^T$ are there?

Question

How many $2 × 2$ matrices $A$ satisfy both $A^3 = I_2$ and $A^ 2 = A^t$ , where $I_2$ denotes the $2 × 2$ identity matrix and $A^ t$ denotes the transpose of $A$?

A bit of manipulation gives me $AA^t=A^tA=I_2$.So this is orthogonal.

Now is there a fixed number of $2\times2$ orthogonal matrices? I have no idea. Please help.

Such a matrix must be a rotation matrix of order $3$, so what angle must the rotation be through? — Angina Seng, May 28 '19 at 15:26
But the answer is infinitely many, since the axis of rotation is arbitrary. — Robert Israel, May 28 '19 at 15:28
The title did not mention the other conditions. Of course, there are infinitely many (real) orthogonal matrices of size $2$. — Dietrich Burde, May 28 '19 at 15:28
@LordSharktheUnknown Should I substitute an arbitrary rotation matrix in the conditions and find $\theta$? — Soham, May 28 '19 at 15:30

score 5 · Accepted Answer · answered May 28 '19 at 15:31

While there are infinitely many $2×2$ orthogonal matrices, there are only two types of them: rotations and reflections. After deriving that $A$ is orthogonal, $A^2=A^T$ is equivalent to $A^2=A^{-1}$ and $A^3=I$, so the two conditions become equivalent.

A non-identity reflectional orthogonal matrix always satisfies $A^2=I$, and so can never satisfy $A^3=I$. Thus $A$ is a rotational matrix, and there are exactly three rotation angles that allow $A^3=I$ to be satisfied: $0$ (the identity), $2\pi/3$ and $-2\pi/3$, yielding the solutions to $A$ as $I_2$, $$B=\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}$$ and $B^{-1}=B^T$.

How many $2\times2$ orthogonal matrices $A$ with $A^3=I$ and $A^2=A^T$ are there?

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