This might be a non-trivial and hard problem. I've been thinking about this for days but couldn't find a good answer, so I hope any of you could give me a good answer/intuition for me to move forward.
Suppose I have two pure states each being $n$ qubits, say $\lvert \psi \rangle$ and $\lvert \phi \rangle$. What is a good (not necessarily the best, but efficient) way to show that $\lvert \psi \rangle$ and $\lvert \phi \rangle$ are locally unitary equivalent, i.e. $\lvert \psi \rangle = (U_1 \otimes U_2 \otimes \cdots \otimes U_n)\lvert \phi \rangle$? Furthermore, given two vector spaces, what is a good way to show these two vector spaces are locally unitary equivalent?
We can perform a partial trace on every subsystem from $1$ to $n$ and find unitaries that map two reduced states of each system, from $U_1$ to $U_N$. Then we can check whether $U = U_1 \otimes U_2 \otimes \cdots \otimes U_n$ indeed maps one state to another. But this is complicated in the case of vector spaces because we need to find proper $U_i$'s such that it maps every orthonormal basis of one vector space to another set of orthonormal basis.
I have another idea that uses the fact for bipartite quantum states, two states are locally unitary equivalent if and only iff Schmidt coefficients between the two states are the same. So we can perform Schmidt decomposition to $n$ qubit states into two parts, say $H_1$ and $H_2\otimes \cdots \otimes H_n$, and check whether two reduced states have the same Schmidt coefficients. Then, we repeat this by Schmidt decomposing $H_i$ and $H_1 \otimes \cdots \otimes H_n$ from $i=2, ..., n$. If, for all cases, Schmidt coefficients are the same, then I think we can conclude that there exists a local unitary.
Update: I believe no matter what methods I use, bi-partitioning should be considered in any aspect.