= Solution
Introduce the orthonormal local vectors $|+\rangle=(|1\rangle+|2\rangle)/\sqrt2$ and $|-\rangle=(|1\rangle-|2\rangle)/\sqrt2$. Reexpressing the second state's factors in this basis gives its <Schmidt decomposition>
$$
|\phi_2\rangle=\frac1{\sqrt3}|+\rangle_A|+\rangle_B+\sqrt{\frac23}|-\rangle_A|-\rangle_B.
$$
The squared <Schmidt coefficients> are therefore $1/3,2/3$, whereas those of the first state are $1/2,1/2$.
To see directly why these distinguish the states, a <local unitary operation> $U_A\otimes U_B$ changes the <reduced density matrix> to
$$
\rho_A' =\operatorname{Tr}_B[(U_A\otimes U_B)\rho_{AB}(U_A^\dagger\otimes U_B^\dagger)]
=U_A\rho_AU_A^\dagger.
$$
This is <local-unitary invariance of reduced-state spectra>: the unitary on the traced subsystem cancels, and conjugation on $A$ leaves its <eigenvalues> unchanged. Here the reductions are
$$
\rho_A^{(1)}=\frac I2,\qquad
\rho_A^{(2)}=\frac13|+\rangle\langle+|+\frac23|-\rangle\langle-|
=\begin{pmatrix}1/2&-1/6\\-1/6&1/2\end{pmatrix}.
$$
Their spectra differ. Hence \b[no pair of local unitary operations carries the first state to the second]. More generally, normalized bipartite pure states are locally unitarily equivalent exactly when their squared Schmidt coefficients agree, including multiplicities; sufficiency follows by mapping one pair of Schmidt bases to the other.
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