Does there or does there not exist $A$ of infinite order in $\{ A \in GL_2(\mathbb{R}) : A^T = A^{-1} \}$?
I know that elements of the form $$A=\begin{bmatrix}\cos(\theta) & \sin(\theta)\\\pm\sin(\theta) & \pm\cos(\theta)\end{bmatrix}$$ are in this group, but do not know how to calculate their order.
If we let $$ A_\theta = \begin{pmatrix}\cos \theta & -\sin\theta\\\sin\theta & \cos \theta\end{pmatrix} $$ be a rotation matrix representing a rotation of the plane by $\theta$, then we have $A_\theta A_\theta^T = A_\theta^TA_\theta = I$, which can be shown by carrying out the multiplications, or by noting that $A^T_\theta = A_{-\theta}$ is the opposite rotation of $A_\theta$. Now if we let $\theta$ be an irrational multiple of $\pi$ (like $1, \pi^2, e\pi, \sqrt 2, \ln 3$, there are many) then no positive power of $A_\theta$ will bring the plane back to where it started, so the order of $A_\theta$ is infinite.