Is $\operatorname{Fun}^\text{small}(\operatorname{Fun}(C,\mathsf{Set}),\mathsf{Set})$ total when $C$ is small? Let $\mathcal C$ be a small category. Then it is known that the category $\operatorname{Fun}^\text{small}(\operatorname{Fun}(\mathcal C,\mathsf{Set}),\mathsf{Set})$ of functors $\operatorname{Fun}(\mathcal C,\mathsf{Set}) \to \mathsf{Set}$ which are small colimits of representable functors is cocomplete and (less obviously) complete. For instance, this is shown in ABLR - A classification of accessible categories, and also in Day and Lack - Limits of small functors.
Question: Is this category total? Is it cototal?
(Totality is a property which is implied by local presentability  and which implies both completeness and cocompleteness. I am fairly certain that the category in question is not locally presentable unless $\mathcal C$ is empty, but if it were total, it would be “almost as good” as being locally presentable.)
One sufficient criterion for totality is for a category to be cocomplete, be epi-complete, and have a generator. So it would be useful to know along the way whether this category has a generator, is cowellpowered, etc.
To simpify notation, write $P(\mathcal C) = \operatorname{Fun}^\text{small}(\mathcal C^{op},Set)$ and $L(\mathcal C) = P(\mathcal C^{op})^{op}$. In these terms, the question asks, for $\mathcal C$ small, whether $P(L(\mathcal C))$ is total.
 A: I think one can see that even when $\mathcal C = 1$, the category $P(L(1)) = Acc(Set)$ of accessible functors $Set \to Set$ is not total or cototal, or even compact or cocompact in the sense of Isbell. In other words, there exist continuous functors $\Phi: Acc(Set)^{op} \to Set, \Psi: Acc(Set) \to Set$ which are not representable:
Proof of Non-cototality / cocompactness:
For sufficiently large regular cardinals $\kappa$, choose some $F_\kappa \in Acc(Set)$ such that $F_\kappa(X) = 1$ for $|X|<\kappa$, but $|F_\kappa(X)| > 1$ for $|X| = \kappa$. For instance, we may take $F_\kappa(X) = X^\kappa / X^{<\kappa}$ to be the set of $\kappa$_sequences in $X$ modulo those which are eventually constant. Then let $\Phi: Acc(Set)^{op} \to Set$ be defined by $\Phi(G) = \prod_\kappa Nat(G,F_\kappa)$. This functor is well-defined because if $G$ is $\lambda$-accessible, then $Nat(G,F_\kappa) = 1$ for $\kappa \geq \lambda$, so that the indicated product is small. Moreover, as a product of continuous functors, this functor is continuous. However, this functor is not representable. If it were, then by testing with the objects $G = Hom(X,-) \in Acc(Set)$, we would see that the representing functor is $\prod_\kappa F_\kappa: Set \to Set$. But this functor is not $\lambda$-accessible for any $\lambda$, since it has as a retract the functor $\prod_{\kappa \geq \lambda} F_\kappa$, which is clearly not $\lambda$-accessible.
Proof of Non-totality / compactness:
As before, set $F_\kappa(X) = X^\kappa / X^{<\kappa}$ for each regular $\kappa$, and let $\Psi(G) = \prod_{\kappa,G(\ast)} Nat(F_\kappa,G)$ to be the large fiber product of the representables on these functors, with the fiber product taken over the functor $ev_\ast$ which evaluates at a point. Note that if $G$ is $\lambda$-accessible, then for $\kappa \geq \lambda$ we have that $G(\kappa) = \varinjlim_{\alpha < \kappa} G(\alpha) \to \varprojlim_{\alpha < \kappa} G(\alpha)$ is injective. It follows that $Nat(F_\kappa, G) \to G(\ast)$ is likewise injective for $\kappa \geq \lambda$. Therefore $\Psi(G)$ is always a small set. Moreover, as a product of representable functors, $\Psi$ is continuous. Further, the unique map $\kappa \to \ast$ induces a map $ev_\ast \to \Psi$, which is a section of the canonical map in the other direction. Using this map, we also see that for each regular $\lambda$, the representable on $F_\lambda$ (which is not $\kappa$-accessible for $\kappa < \lambda$) is a retract of $\Psi$. Thus $\Psi$ is not accessible, and therefore not corepresentable.
