How can I functorially dualise in a symmetric monoidal $(\infty,1)$-category with duals?

One way to construct the duality functor ${\cal C} \to {\cal C^{\rm op}}$ is through the notion of a pairing of $\infty$-categories (see HA, Definition 5.2.1.5). In particular, in this case we're talking about a self-pairing on ${\cal C}$, which by definition is a right fibration $\mu:{\cal M} \to {\cal C} \times {\cal C}$, classified by a functor $b: {\cal C}^{\rm op} \times {\cal C}^{\rm op} \to {\cal S}$ to spaces. We say that $\mu$ is left representable if for every $x \in {\cal C}$ the functor $y \mapsto b(x,y)$ is representable in ${\cal C}$, and right representable if for every $y \in {\cal C}$ the functor $x \mapsto b(x,y)$ is represetable in ${\cal C}$. If $\mu$ is a pairing which is both left and right representable then it determines an adjunction between ${\cal C}$ and ${\cal C}^{\rm op}$. We say that $\mu$ is a perfect pairing if it is both left and right representable and the associated adjunction is an equivalence. One can then show that the data of a perfect pairing $\mu: {\cal M} \to {\cal C} \times {\cal C}$ is in fact equivalent to the data of an equivalence ${\cal C} \to {\cal C}^{\rm op}$. The advantage of this point of view is that the $\mathbb{Z}/2$-action on the space of equivalences ${\cal C} \to {\cal C}^{\rm op}$ becomes explicit: it is given by sending $\mu: {\cal M} \to {\cal C} \times {\cal C}$ to ${\rm swap} \circ \mu: {\cal M} \to {\cal C} \times {\cal C}$, where ${\rm swap}:{\cal C} \times{\cal C} \to {\cal C} \times {\cal C}$ is the equivalence that swaps the two components. In particular, to construct an equivalence ${\cal C} \stackrel{\simeq}{\to} {\cal C}^{\rm op}$ which is self-dual in a homotopy coherent manner is equivalent to constructing a perfect pairing ${\cal M} \to {\cal C} \times {\cal C}$ together with a $\mathbb{Z}/2$-action on ${\cal M}$ which lifts the swap action on ${\cal C} \times {\cal C}$. Note that this swap action can itself be encoded as a right fibration $p:{\rm Sym}({\cal C}) \to {\rm B}(\mathbb{Z}/2)$, where ${\rm B}(\mathbb{Z}/2)$ is the groupoid with one object whose endomorphism group is $\mathbb{Z}/2$, and the fiber of $p$ over this one object is ${\cal C} \times {\cal C}$. One can then encode a lift of the swap action to ${\cal M}$ by a right fibration $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ whose restriction to ${\cal C} \times {\cal C}$ is ${\cal M}$. In general, we may refer to right fibrations $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ as symmetric pairings, and say that a symmetric pairing is perfect when its base change to ${\cal C} \times {\cal C} \subseteq {\rm Sym}({\cal C})$ is perfect. The notion of a perfect symmetric pairing then encodes a duality on ${\cal C}$, i.e., a self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$ (with all the higher coherences taken into account).

Now in the case of a symmetric monoidal $\infty$-category with duals, the pairing encoding the duality is the right fibration ${\cal M} \to {\cal C} \times {\cal C}$ classified by the functor $b_{\cal C}: {\cal C} \times {\cal C} \to {\cal S}$ sending $(x,y)$ to ${\rm Map}_{\cal C}(x \otimes y,1_{\cal C})$. There is a direct way to construct this as a symmetric perfect pairing. Indeed, suppose that $\pi:{\cal C}^{\otimes} \to {\rm Fin}_*$ is the coCartesian fibration encoding the symmetric monoidal structure on ${\cal C}$, and let ${\cal C}^{\otimes}_{\rm act} \to {\rm Fin}$ be its active part, i.e., the base change to the category ${\rm Fin}$ of finite sets via the functor $I \mapsto I_+$. Then we can identify ${\rm B}(\mathbb{Z}/2)$ as the subcategory of ${\rm Fin}$ consisting of the object $\left<2\right>^{\circ} = \{1,2\} \in {\rm Fin}$ and all its automorphisms, and the base change of ${\cal C}^{\otimes}_{\rm act}$ to ${\rm B}(\mathbb{Z}/2) \subseteq {\rm Fin}$ is naturally equivalent to ${\rm Sym}({\cal C})$. If $1_{\cal C} \in ({\cal C}^{\otimes}_{\rm act})_{\left<1\right>^{\circ}}$ now denotes the unit then the right fibration $$ ({\cal C}^{\otimes}_{\rm act})_{/1_{\cal C}}\times_{{\rm Fin}} {\rm B}(\mathbb{Z}/2) \to {\cal C}^{\otimes}_{\rm act} \times_{{\rm Fin}}{\rm B}(\mathbb{Z}/2) \simeq {\rm Sym}({\cal C}) $$ is a symmetric pairing whose underlying pairing is $b_{\cal C}$ above. When ${\cal C}$ has duals this symmetric pairing is perfect, yielding the associated self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$.