Left and right duals, and double duals, need not collapse

Let $\mathbb{Z}$ be the ordered set of integers regarded as a small category, so there is a unique morphism $m\to n$ exactly when $m\le n$. Let $D(n)=2n$, $L(n)=\lceil n\mathrm{/}2\rceil$, $R(n)=\lfloor n\mathrm{/}2\rfloor$, and $H(n)=2n-1$. Directly from the defining inequalities, $L(m)\le n⟺m\le D(n),D(m)\le n⟺m\le R(n),H(m)\le n⟺m\le L(n)\mathrm{.}$ Thus $L⊣D⊣R$ and $H⊣L$.

Let $\mathcal{R}$ be the full subcategory of $\mathrm{End}(\mathbb{Z})$ generated under finite composition by the identity functor and all endofunctors in the bi-infinite adjoint chain obtained by repeatedly taking left and right adjoints of $D$. It is a monoidal subcategory by construction. Each generator has both adjacent adjoints in the chain, and a finite composite of functors with left and right adjoints has the corresponding reversed composites as its left and right adjoints. Those composites again belong to $\mathcal{R}$, and their units and counits lie in the full subcategory. Hence every object of $\mathcal{R}$ has both duals and $\mathcal{R}$ is rigid.

In the endofunctor category of a poset, a natural transformation $F\Rightarrow G$ exists exactly when $F(n)\le G(n)$ for all $n$, so two endofunctors are isomorphic exactly when they are equal pointwise. Here $L(1)=1>0=R(1)$, so there is no natural transformation $L\Rightarrow R$ and therefore no isomorphism $L\cong R$. Thus the object $D$ has nonisomorphic left and right duals.

If the chosen left duality on $\mathcal{R}$ uses left adjoints, then $D^{\vee }=L$ and $D^{\vee \vee }=H$. Since $H(0)=-1\mathrm{\ne }0=D(0)$, the endofunctors $H$ and $D$ are not isomorphic. Therefore in this rigid monoidal category neither left/right duals nor double duals are forced to collapse.