Kan extensions are unique up to unique isomorphism

Since $(L,\eta )$ is a left Kan extension and ${\eta }^{\mathrm{\prime }}:F\Rightarrow L^{\mathrm{\prime }}K$ is another such pair, [L1] gives a unique natural transformation $\alpha :L\Rightarrow L^{\mathrm{\prime }}$ with ${\eta }^{\mathrm{\prime }}=(\alpha K)\circ \eta$; similarly [L1] gives a unique natural transformation ${\alpha }^{\mathrm{\prime }}:L^{\mathrm{\prime }}\Rightarrow L$ with $\eta =({\alpha }^{\mathrm{\prime }}K)\circ {\eta }^{\mathrm{\prime }}$.

By step 1.1, $(({\alpha }^{\mathrm{\prime }}\alpha )K)\circ \eta =({\alpha }^{\mathrm{\prime }}K)\circ {\eta }^{\mathrm{\prime }}=\eta$, while $(1_{L}K)\circ \eta =\eta$ trivially. So the uniqueness clause of [L1] forces ${\alpha }^{\mathrm{\prime }}\alpha =1_L$; likewise $\alpha {\alpha }^{\mathrm{\prime }}=1_{L^{\mathrm{\prime }}}$. Hence $\alpha$ is a natural isomorphism, and its compatibility with $\eta$ was built in at step 1.1.

The same argument with the terminal clause of [L1] gives unique $\beta :R\Rightarrow R^{\mathrm{\prime }}$ and ${\beta }^{\mathrm{\prime }}:R^{\mathrm{\prime }}\Rightarrow R$ satisfying $\epsilon ={\epsilon }^{\mathrm{\prime }}\circ (\beta K)$ and ${\epsilon }^{\mathrm{\prime }}=\epsilon \circ ({\beta }^{\mathrm{\prime }}K)$, and uniqueness forces ${\beta }^{\mathrm{\prime }}\beta =1_R$ and $\beta {\beta }^{\mathrm{\prime }}=1_{R^{\mathrm{\prime }}}$. So right Kan extensions are unique up to unique compatible isomorphism as well.