Suppose I have a symmetric square matrix $\mathbb A$ in block-form:
$$\mathbb{A}= \left(\begin{array}{cc} \mathbb{A}_{11} & \mathbb{A}_{12}\\ \mathbb{A}_{21} & \mathbb{A}_{22} \end{array}\right)$$
where $\mathbb{A}_{21} = \mathbb{A}_{12}^T$ and $\mathbb A_{11}$ and $\mathbb A_{22}$ are symmetric.
Now suppose I apply a similarity transformation that acts block-wise, that is:
$$\mathbb{A}' = \left(\begin{array}{cc} \mathbb{P}\mathbb{A}_{11} \mathbb{P}^{- 1} & \mathbb{P}\mathbb{A}_{12} \mathbb{Q}^{- 1}\\ \mathbb{Q}\mathbb{A}_{21} \mathbb{P}^{- 1} & \mathbb{Q}\mathbb{A}_{22} \mathbb{Q}^{- 1} \end{array}\right)$$
where $\mathbb P$ and $\mathbb Q$ are suitable invertible matrices.
Finally I invert $\mathbb A'$, and express the inverse block-form using the same layout:
$$(\mathbb{A}')^{-1} = \left(\begin{array}{cc} \mathbb{B}_{11}' & \mathbb{B}_{12}'\\ \mathbb{B}_{21}' & \mathbb{B}_{22}' \end{array}\right)$$
Now my question is this: Can I always choose $\mathbb P$ and $\mathbb Q$ such that $\mathbb B_{11}'$ and $\mathbb B_{22}'$ are diagonal? If the answer is yes, can we also chose them to be orthogonal?
Yes, assuming you're talking about real matrices. The Spectral Theorem tells you that any symmetric real matrix can be diagonalized by an orthogonal change of basis.