= Solution
For $b:B\to B'$ in $\mathcal B$, conjugate the induced <natural transformation> by the chosen <representations of a functor>:
$$
\tau_b=\psi_{B'}^{-1}\,H(-,b)\,\psi_B:
\mathcal A(-,GB)\Rightarrow\mathcal A(-,GB').
$$
The <Yoneda lemma> gives a unique <morphism> $Gb:GB\to GB'$ whose postcomposition map is $\tau_b$. Explicitly,
$$
\boxed{Gb=(\psi_{B',GB})^{-1}\!\left(H(GB,b)(\psi_{B,GB}(1_{GB}))\right).}
$$
Consequently, for every $f:A\to GB$,
$$
H(A,b)(\psi_{B,A}(f))=\psi_{B',A}(Gb\circ f).
$$
This is precisely the required <naturality> in $B$. The identities $\tau_{1_B}=1$ and $\tau_{b'b}=\tau_{b'}\tau_b$ force $G1_B=1_{GB}$ and $G(b'b)=Gb'\,Gb$, because the <Yoneda embedding> is <fully faithful>. Thus \b[the representing objects extend uniquely to a functor] $G:\mathcal B\to\mathcal A$. Uniqueness follows by the same <Yoneda lemma>: any candidate satisfying that <naturality> must induce $\tau_b$, so must have exactly the displayed action on <morphisms>.
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