= Solution
Suppose first that the stated unitary and environment states exist. Preservation of the <inner product> gives
$$
\langle\phi_0|\phi_1\rangle
=\langle\psi_0|\psi_1\rangle
\langle e_0|e_1\rangle.
$$
The environment states are normalized, so the <Cauchy-Schwarz inequality> gives $|\langle e_0|e_1\rangle|\leq1$. Therefore
$$
|\langle\phi_0|\phi_1\rangle|
\leq|\langle\psi_0|\psi_1\rangle|.
$$
Conversely, write
$$
a=\langle\phi_0|\phi_1\rangle,
\qquad b=\langle\psi_0|\psi_1\rangle
$$
and assume $|a|\leq|b|$. If $b\ne0$, put $c=a/b$, so $|c|\leq1$, and choose
$$
|e_0\rangle=|0\rangle,
\qquad
|e_1\rangle=c|0\rangle+\sqrt{1-|c|^2}\,|1\rangle.
$$
Then $\langle e_0|e_1\rangle=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0\rangle=|e_1\rangle=|0\rangle$.
In either case the input pair and output pair have the same <Gram matrix>. The isometry taking one pair to the other extends to a <unitary matrix>, so the required $U$ exists. Hence
$$
\boxed{
U\text{ exists}
\quad\Longleftrightarrow\quad
|\langle\phi_0|\phi_1\rangle|
\leq|\langle\psi_0|\psi_1\rangle|
},
$$
which is the <environment-assisted two-state pure-state transformation> criterion.
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