Adjoints are unique up to a unique natural isomorphism compatible with the adjunction data

By initiality of $(Fc,{\eta }_c)$ there is a unique ${\alpha }_c:Fc\to F^{\mathrm{\prime }}c$ with $G({\alpha }_c){\eta }_c={\eta }_c^{\mathrm{\prime }}$; reversing the roles gives a unique ${\beta }_c:F^{\mathrm{\prime }}c\to Fc$ with $G({\beta }_c){\eta }_c^{\mathrm{\prime }}={\eta }_c$.

The composites ${\beta }_c{\alpha }_c$ and $1_{Fc}$ both carry ${\eta }_c$ to itself, so initiality gives ${\beta }_c{\alpha }_c=1_{Fc}$; similarly ${\alpha }_c{\beta }_c=1_{F^{\mathrm{\prime }}c}$.

Let $a:c\to c^{\mathrm{\prime }}$. The object $(F^{\mathrm{\prime }}{c}^{\mathrm{\prime }},{\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}\circ a)$ lies in $(c\downarrow G)$, so by initiality of $(Fc,{\eta }_c)$ exactly one morphism $Fc\to F^{\mathrm{\prime }}{c}^{\mathrm{\prime }}$ composes with ${\eta }_c$ to ${\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$. Both candidates do: $G(F^{\mathrm{\prime }}a{\alpha }_c){\eta }_c=G(F^{\mathrm{\prime }}a){\eta }_c^{\mathrm{\prime }}={\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$ by naturality of ${\eta }^{\mathrm{\prime }}$, and $G({\alpha }_{c^{\mathrm{\prime }}}Fa){\eta }_c=G({\alpha }_{c^{\mathrm{\prime }}}){\eta }_{c^{\mathrm{\prime }}}a={\eta }_{c^{\mathrm{\prime }}}^{\mathrm{\prime }}a$ by naturality of $\eta$. Hence $F^{\mathrm{\prime }}a{\alpha }_c={\alpha }_{c^{\mathrm{\prime }}}Fa$ and $\alpha$ is natural. This uses only initiality and naturality of the units, so no local smallness is required.

Any natural transformation compatible with the units has components satisfying the uniqueness condition in step 1.1 and hence equals $\alpha$. For counit compatibility, both ${\epsilon }_d$ and ${\epsilon }_d^{\mathrm{\prime }}\circ {\alpha }_{Gd}$ are morphisms $FGd\to d$, and initiality of $(FGd,{\eta }_{Gd})$ determines such a morphism by its composite: $G({\epsilon }_d){\eta }_{Gd}=1_{Gd}$ by the triangle identity of [L1], while $G({\epsilon }_d^{\mathrm{\prime }})G({\alpha }_{Gd}){\eta }_{Gd}=G({\epsilon }_d^{\mathrm{\prime }}){\eta }_{Gd}^{\mathrm{\prime }}=1_{Gd}$ by step 1.1 and the triangle identity for $F^{\mathrm{\prime }}⊣G$. So ${\epsilon }_d^{\mathrm{\prime }}\circ {\alpha }_{Gd}={\epsilon }_d$. Passing to opposite categories proves the dual assertion.