Let $\mathbf{A} \in \mathbb{C}^{N \times N}$ be a Hermitian matrix and let $\mathbf{D} \in \mathbb{C}^{N \times K}$ (with $K<N$) be a semi-unitary matrix such that $\mathbf{D}^{H} \mathbf{D} = \mathbf{I}_{K}$.
What properties does $\mathbf{A}$ have if the following relationship is satisfied?
\begin{align} \mathbf{D}^H \mathbf{A}^{-1} \mathbf{D} & = \mathbf{D}^H \mathbf{D}(\mathbf{D}^H \mathbf{A} \mathbf{D})^{-1} \mathbf{D}^H \mathbf{D} \\ & = (\mathbf{D}^H \mathbf{A} \mathbf{D})^{-1} \in \mathbb{C}^{K \times K} \end{align}
Note that $\mathbf{D}^H \mathbf{A} \mathbf{D}$ is not a diagonal matrix. Also, note that the above relationship is always satisfied when $K=N$, i.e., $\mathbf{D} \in \mathbb{C}^{N \times N}$ is a unitary matrix.
By a change of orthonormal basis you may assume that $D=\pmatrix{I_K\\ 0}$. Partition $A$ as $\pmatrix{X&Y^H\\ Y&Z}$, where $X$ is $K\times K$. The given condition implies that $X$ is invertible. Since $A$ is also invertible, the Schur complement $S=Z-YX^{-1}Y^H$ is invertible and $$ A^{-1}=\pmatrix{X^{-1}+X^{-1}Y^HS^{-1}YX^{-1}&\ast\\ \ast&S^{-1}}. $$ The condition $(D^HAD)^{-1}=D^HA^{-1}D$ thus implies that $X^{-1}=X^{-1}+X^{-1}Y^HS^{-1}YX^{-1}$, i.e., $Y^HS^{-1}Y=0$. I don't think this has much significance in general.
However, if $A$ happens to be positive definite, then $S$ is necessarily positive definite. So, from $Y^HS^{-1}Y=0$ we obtain $Y=0$ and in turn $A=X\oplus Z$. Therefore, in this case, the range of $D$ and its orthogonal complement are invariant subspaces of $A$.