Graduate Is the 2-qubit singlet state invariant under bilateral projections?

[QuantumExplorer]

The 2-qubit singlet state $\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{01}-\ket{10})$ is invariant under bilateral rotations(https://arxiv.org/pdf/quant-ph/9511027), that is any operator of the form $U \otimes U$ for $U$ being unitary. However, I'm wondering if the state is invariant also for a more general transformation $A \otimes A$. In general that transformation will lead to $(A \otimes A )\ket{\Psi^-} = \det(A)\ket{\Psi^-}$. Given this, would it be OK to say that $\ket{\Psi^-}$ is invariant for any $A$, or do we have to worry about the cases when $A$ is not invertible? If $A$ is singular, say $A = \ket{0}\bra{0}$, then in this case $\det(A)=0$ and the application of $A \otimes A$ will result in null vector, which seems different enough from the original to still call the case "invariant".