= Solution
Assume the <hom-set> condition. Its surjectivity at $B'=B$ gives $u:KB\to B$ with $u\rho_B=1_B$. We show that this splitting is two-sided.
Because $J$ is <fully faithful>, $\varepsilon_{KB}$ is invertible. Both <triangle identities for an adjunction> give, after suppressing $J$,
$$
K\rho_B=\rho_{KB}=\varepsilon_{KB}^{-1}.
$$
Applying $K$ to $u\rho_B=1_B$ yields $Ku\,\rho_{KB}=1_{KB}$. <Naturality> of the <adjunction unit> at $u$ gives $\rho_Bu=Ku\,\rho_{KB}$, hence $\rho_Bu=1_{KB}$. Consequently \b[the hom-set condition forces the unit to be invertible]:
$$
\boxed{\rho_B^{-1}=u.}
$$
Only surjectivity at $B'=B$ was needed for this direction; the given family of <bijections> certainly supplies it.
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