Dur, 2000 states that
(...)But even in the simplest systems, $|\psi\rangle$ and $|\phi\rangle$ are typically not related by LU, and continuous parameters are needed to label all equivalence classes.
I've found some similar explanation in Ritz,2018
(...)The Schmidt decomposition of a two-qubit state has only one free parameter $$|\psi\rangle = \sqrt{\lambda_{0}}|00\rangle+\sqrt{\lambda_{1}}|11\rangle \quad ;\lambda_{0}+\lambda_{1}=1 \qquad\tag{2.85}$$ Thus, we can rewrite eq. (2.85) in terms of new parameter $\theta$ as $$|\psi\rangle = \cos \theta |00\rangle+\sin \theta|11\rangle \qquad \qquad \qquad \qquad \quad\tag{2.86}$$ Therefore, any two-qubit state can, under LU, be transformed to Eq. (2.86). Obviously there is still one continuous parameter, i.e. $θ$, left. Hence, even for the lowest possible dimension and particles, the number of equivalence classes under LU is infinite.
I couldn't understand, why any abritary two-qubit states under LU can be transformed to eq. (2.86)? If it's LU transformation, I think we should start from the general form of Unitary matrices itself, but it seems that we choose convinient transformation such that $\lambda_0 = \cos^2 \theta$ and $\lambda_1 = \sin^2 \theta$. If that so, why does continuous parameter make the number of classes infinte? I feel kinda clueless here.