As discussed in What is the geometrical action of a skew-symmetric matrix on an arbitrary vector?, an arbitrary real skew-symmetric matrix can be brought into the block diagonal form $$\begin{bmatrix} \begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} & 0 & \cdots & 0 \\ 0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \\ & & & & \begin{matrix}0 \\ & \ddots \\ & & 0 \end{matrix} \end{bmatrix}$$ via a real orthonormal similarity transformation. This demonstrates that the operator's geometric action is to annihilate components of vectors in certain directions and dilate and rotate other components by $\pi/2$ in the $x^i-x^{i+1}$ planes in other directions.
But I'm curious what happens in the "degenerate" case. Without significant loss of generality, let's consider the simplest case where we have a single pair of degenerate blocks; i.e. the operator $A$ is orthonormally similar to $$\left[ \begin{matrix} 0 & \lambda & 0 & 0 \\ -\lambda & 0 & 0 & 0 \\ 0 & 0 & 0 & \lambda \\ 0 & 0 & -\lambda & 0 \end{matrix} \right].$$
Again without significant loss of generality, we can further simplify by rescaling the operator by $1/\lambda$ to get that $A$ is (up to an overall constant) orthonormally similar to $$\left[ \begin{matrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{matrix} \right].$$
Is there a simple geometric characterization of the set of such operators $A$ (beyond the generic situation described above)?
Roughly speaking, I'm curious about the geometric symmetry group of such an operator. For example: based on my experience with diagonalizable matrices, I suspect that such a "degenerate" operator may be secretly isotropic (or at least have a higher degree of geometric symmetry than the generic antisymmetric case). Is this the case, or does there exist a unique pair of orthogonal planes in $\mathbb{R}^4$ about which these rotations occur? If it is isotropic, is there a simpler/more geometrically intuitive description of its action, which manifestly demonstrates the isotropy?
