Let me try to summarize my question by taking into consideration some elementary notions of orthogonal matrices.
The general $2\times2$ orthogonal matrix has the form:
$$O = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} $$
Where $\theta$ is an arbitrary real number. It follows that $O^T O = I$. This matrix is also capable of diagonalizing some symmetric matrix $A$:
$$O^T A O = \text{diag}(A)$$
Now, let's assume that $\theta$ is not a real number, but rather, is a matrix. Therefore:
$$O^T O = \begin{pmatrix}\cos\theta^T \cos\theta + \sin\theta^T \sin\theta & \cos\theta^T\sin\theta - \sin\theta^T\cos\theta \\ -\sin\theta^T\cos\theta + \cos\theta^T\sin\theta & \cos\theta^T\cos\theta + \sin\theta^T\sin\theta \end{pmatrix}$$
It is no longer clear that $O^T O$ is the identity. It actually seems like:
$$O^\prime = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
is the inverse of $O$ instead of $O^T$. Moreover, there also seems to be an issue with the diagonalization of our matrix $A$, it seems like $O^T A O$ would no longer yield a diagonal matrix because $O$ and $O^T$ are no longer the inverse of each other (maybe I am wrong with my reasoning?), but if I use $O^\prime A O$ instead, there is still an issue since it would seem as if the diagonal matrix is not symmetric, and, therefore, can't be diagonal.
Is this generalization completely nonsensical or am I missing something here?