= Solution
Share the <Bell state> $|\Phi_+\rangle$ between Alice and Bob, and allow their actions to be spacelike separated. The <projective measurement> postulate gives the <remote preparation in conjugate bases> identity
$$
({}_A\langle a|\otimes I_B)|\Phi_+\rangle
=\frac{|a^*\rangle_B}{\sqrt2}.
$$
If Alice measures in the computational basis, Bob is prepared in $|0\rangle$ or $|1\rangle$, each with probability one half. If she instead measures in the conjugate basis $U^*|0\rangle,U^*|1\rangle$, Bob is prepared in $U|0\rangle$ or $U|1\rangle$, again equally likely. The conjugation is essential for a general complex <unitary operator>.
Interpreting distinct states as distinct physical rays, the two candidate pairs are disjoint. A hypothetical perfect identifier would tell Bob which member he received and therefore which basis Alice chose. Alice could transmit a bit by that basis choice, without sending her outcome or any other message. Bob could read it while spacelike separated from Alice, violating the no-superluminal-signalling requirement of <special relativity>. This establishes that <perfect two-basis discrimination would allow superluminal signalling> and rules out the device using the requested physical assumptions.
Ordinary <quantum measurements> do not signal: the two equally weighted ensembles have identical <reduced density matrix>
$$
\frac12\sum_{j=0}^1|j\rangle\langle j|
=\frac12\sum_{j=0}^1U|j\rangle\langle j|U^\dagger
=\frac I2.
$$
The contradiction arises from the hypothesized ability to label the ensemble's individual pure members. Global-phase variants of a vector do not count as distinct physical <pure states>; treating them as distinguishable labels would already introduce an unphysical extra readout.
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