Given an invertible matrix $A \in \mathbb{R}^{n \times n}$, and index set $\langle n\rangle = \{ 1, \dots, n \}$, and the submatrix $A(\alpha)$ with the columns and rows of $A$ with indices $\alpha \subset \langle n \rangle$.
Is there are characterisation of all matrices $A$ with
$\textrm{det } A(\alpha) = \textrm{det } A^{-1}(\alpha) \qquad \forall \alpha \subset \langle n \rangle \qquad ?$
Example: Sufficient conditions are involutory ($A^{-1} = A$) or orthogonal ($A^{-1} = A^T$) matrices.
Let $A=(a_{ij})$. Then we obtain a system of polynomial equations in the variables $a_{ij}$, given by determinant conditions $\det A(\alpha)=\det A^{-1}(\alpha)$. If one determines a Groebner basis for such a system with a given $n$, we see that the dimension of the variety given by the polynomial equations is rather high. We see that there are always many other solutions for $A$, which do not satisfy $A=A^{-1}$ or $A^{-1}=A^T$. One reason is, that many determinants of the submatrices of size less than $n$ can be chosen to be zero, for $A$ and $A^{-1}$.
I know this does not answer the question how to give a nice characterization for this variety other than by determinants. Therefore, let me just give a typical example for $n=3$, with $a,b\in \mathbb{R}$ and $b\neq 0$:
$$ A=\begin{pmatrix} 0 & 0 & -1/b \cr a & 1 & 0 \cr b & 0 & 0 \end{pmatrix}, \quad A^{-1}=\begin{pmatrix} 0 & 0 & 1/b \cr 0 & 1 & -a/b \cr -b & 0 & 0 \end{pmatrix}. $$
