 Solution

Since $a^2=I$, a representation satisfies $D(a{)}^2=I_2$. Orthogonality gives $D(a{)}^{-1}=D(a{)}^T$, while the involution relation gives $D(a{)}^{-1}=D(a)$. Hence $D(a)$ is a symmetric matrix, so $\gamma =\delta$.

Its entries obeyIf $\gamma \ne 0$, then $\beta =-\alpha$; putting $\alpha =\cos \theta$ and $\gamma =\sin \theta$ gives a reflection. If $\gamma =0$, the diagonal entries are independently $\pm 1$: the opposite-sign cases are already included among these reflections, while equal signs give $I_2$ or $-I_2$. Faithfulness excludes $I_2$, since $a\ne I$. Therefore the orthogonal involutions of the plane giveThe first is rotation by $\pi$. The second is reflection in the line making angle $\theta /2$ with the first axis: the unit vector along that line is fixed, and its perpendicular is negated. These are all the nonidentity orthogonal transformations of order two in the plane.