A family into a parametrised end is natural, or dinatural, in the parameter exactly when its composite with the counit is

Fix $u:p\to p^{\mathrm{\prime }}$ and an object $c$ of $\mathcal{C}$. The two morphisms $X(p)\to E(p^{\mathrm{\prime }})$ at issue for the natural clause are ${\varphi }_{p^{\mathrm{\prime }}}\circ X(u)$ and $E(u)\circ {\varphi }_p$, and their composites with ${\omega }_c^{p^{\mathrm{\prime }}}$ are $({\omega }_c^{p^{\mathrm{\prime }}}{\varphi }_{p^{\mathrm{\prime }}})\circ X(u)$ and, by the defining equation of [L1], $T(u,1_c,1_c)\circ ({\omega }_c^p{\varphi }_p)$. Two morphisms $Z\to E(p^{\mathrm{\prime }})$ whose composites with ${\omega }_c^{p^{\mathrm{\prime }}}$ agree for every $c$ are equal, because the common composite family is the terminal wedge precomposed with a morphism, hence a wedge, and it has exactly one factorisation through $E(p^{\mathrm{\prime }})$.

For the forward direction of the natural clause, suppose $\varphi$ is natural in $p$. Then for every $c$ the family ${\omega }_c^{\bullet }{\varphi }_{\bullet }$ is the composite of the natural family $\varphi$ with the family ${\omega }_c^{\bullet }$, which is natural by [L1]; a composite of two natural families is natural by the naturality equation of [F3] applied twice, so ${\omega }_c^{\bullet }{\varphi }_{\bullet }$ is natural in $p$.

For the converse direction of the natural clause, suppose every ${\omega }_c^{\bullet }{\varphi }_{\bullet }$ is natural in $p$. Its naturality equation at $u$ reads $T(u,1_c,1_c)\circ ({\omega }_c^p{\varphi }_p)=({\omega }_c^{p^{\mathrm{\prime }}}{\varphi }_{p^{\mathrm{\prime }}})\circ X(u)$, so by step 1.1 the morphisms ${\varphi }_{p^{\mathrm{\prime }}}X(u)$ and $E(u){\varphi }_p$ have the same composite with ${\omega }_c^{p^{\mathrm{\prime }}}$ for every $c$ and are therefore equal. Since $u$ was arbitrary, $\varphi$ is natural in the parameter.

For the dinatural clause, fix $u:q\to q^{\mathrm{\prime }}$ in $\mathcal{Q}$. The wedge equation for $\psi$ at $u$ is $E(1_q,u)\circ {\psi }_q=E(u,1_{q^{\mathrm{\prime }}})\circ {\psi }_{q^{\mathrm{\prime }}}$, an equation between morphisms $Y\to E(q,q^{\mathrm{\prime }})$, where $(1_q,u)$ and $(u,1_{q^{\mathrm{\prime }}})$ are the two morphisms of ${\mathcal{Q}}^{\mathrm{o}\mathrm{p}}\times \mathcal{Q}$ that the wedge equation names. Composing with ${\omega }_c^{(q,q^{\mathrm{\prime }})}$ and applying the defining equation of [L1] at each of them turns the two sides into $T(1_q,u,1_c,1_c)\circ ({\omega }_c^{(q,q)}{\psi }_q)$ and $T(u,1_{q^{\mathrm{\prime }}},1_c,1_c)\circ ({\omega }_c^{(q^{\mathrm{\prime }},q^{\mathrm{\prime }})}{\psi }_{q^{\mathrm{\prime }}})$, whose equality for every $u$ is exactly the wedge equation for the family ${\omega }_c^{(\bullet ,\bullet )}{\psi }_{\bullet }$ over $T(-,-,c,c)$. So the wedge equation for $\psi$ implies the one downstairs by composition, and conversely if it holds downstairs for every $c$ then the two morphisms of the wedge equation for $\psi$ agree after composition with every ${\omega }_c^{(q,q^{\mathrm{\prime }})}$, hence are equal by step 1.1.