Let me give more details about the comments under Todd's answer.

As Tom pointed out, taking fixed points of the Isbell conjugation for the category $\mathbb{C}$: $$\overset{A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B}{}$$ consisting of two parallel arrows between two different objects, freely adjoins an initial and terminal objects, yielding the category $\mathit{DM}(\mathbb{C})$: $$\overset{0\longrightarrow A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B\longrightarrow 1}{}$$ which is neither complete nor cocomplete (in fact it does not have even binary (co)products).

To see this, let us observe that $\mathbb{C}$ is self-dual, therefore the categories of presheaves and coprseheaves over $\mathbb{C}$ are isomorphic. Both of these categories may be thought of as categories of directed graphs and graph homomorphisms. Let me compute the left part ($L$ in Todd's answer) of the Isbell conjugation $(-)^\star \colon \mathbf{Set}^{\mathbb{C}^{op}} \to (\mathbf{Set}^\mathbb{C})^{op}$: $$G^\star(Y) = \int_{X \in \mathbb{C}} \hom(X, Y)^{G(X)} = \mathit{nat}(G, \hom(-, y))$$ The graph associated to the presheaf $\hom(-, A)$ is: $$\bullet$$ so the set of edges $G^\star(A)$ of $G^\star$ is the set of homomorphisms from graph $G$ to graph $\bullet$. This set is either empty $\emptyset$ --- if $G$ contains any edge, or a singleton $1$ otherwise.

Similarly, the graph associated to the presheaf $\hom(-, B)$ is: $$ \bullet \rightarrow \bullet $$ and so the set of nodes $G^\star(B)$ of $G^\star$ is the set of homomorphisms from graph $G$ to graph $\bullet \rightarrow \bullet$. This set is empty iff $G$ contains composable edges (loops $\looparrowright$ or sequences $\bullet \rightarrow \bullet \rightarrow \bullet$). If $G$ does not contain any sequence of composable edges, then the set of nodes is equal to $2^D$, where $D$ is the largest discrete full subgraph of $G$.

If there is an edge in $G^\star(A)$, and there are at least two distinct nodes in $ G^\star(B)$ then the edge is non-degenerated (its source is different from its target).

By duality, the right part ($R$ in Todd's answer) of the Isbell conjugation $(\mathbf{Set}^\mathbb{C})^{op} \to \mathbf{Set}^{\mathbb{C}^{op}}$ is defined on graphs in essentially the same way.

Here are some examples of graphs $G$ with their corresponding graphs $G^\star$: \begin{array}{ccc} \emptyset & \mapsto & \looparrowright \newline \bullet & \mapsto & \bullet \rightarrow \bullet \newline \looparrowright & \mapsto & \emptyset \newline \bullet \rightarrow \bullet & \mapsto & \bullet \newline \bullet \rightarrow \bullet \; \bullet & \mapsto & \bullet \; \bullet \newline \bullet \; \bullet & \mapsto & \bullet \rightarrow \bullet \; \bullet \; \bullet \newline \bullet \rightarrow \bullet \; \bullet \; \bullet & \mapsto & \bullet \; \bullet \; \bullet \; \bullet \newline \end{array}

One may easily verify that, due to size issues (i.e. there are no isomorphism $2^K \approx K$ for any $K$), only the first four graphs from the above table are fixed-points of the Isbell conjugation. Moreover, these graphs together with their homomorphism constitute a category isomorphic to: $$\overset{0\longrightarrow A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B\longrightarrow 1}{}$$

The gap in Todd's proof of (co)completeness is in the implicit assumption that the fixed-points of a monad are the same as fixed points of the associated idempotent monad. There is an abstract counterexample to this claim --- fixed-points of an idempotent monad constitute a reflective subcategory of the base category, whereas fixed-points of an arbitrary monad generally not (perhaps one of the idea of an associated idempotent monad is just to add "enough" fixed-points to the monad to make the subcategory of fixed-points reflective).

Let me give more details about the comments under Todd's answer.

As Tom pointed out, taking fixed points of the Isbell conjugation for the category $\mathbb{C}$: $$\overset{A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B}{}$$ consisting of two parallel arrows between two different objects, freely adjoins an initial and terminal objects, yielding the category $\mathit{DM}(\mathbb{C})$: $$\overset{0\longrightarrow A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B\longrightarrow 1}{}$$ which is neither complete nor cocomplete (in fact it does not have even binary (co)products).

To see why $\mathit{DM}(\mathbb{C})$ is of the above form, let us first observe that $\mathbb{C}$ is self-dual, therefore the categories of presheaves and coprseheaves over $\mathbb{C}$ are isomorphic. Both of these categories may be thought of as categories of directed graphs and graph homomorphisms. Let me compute the left part ($L$ in Todd's answer) of the Isbell conjugation $(-)^\star \colon \mathbf{Set}^{\mathbb{C}^{op}} \to (\mathbf{Set}^\mathbb{C})^{op}$: $$G^\star(Y) = \int_{X \in \mathbb{C}} \hom(X, Y)^{G(X)} = \mathit{nat}(G, \hom(-, Y))$$ The graph associated to the presheaf $\hom(-, A)$ consitsts of a single node: $$\bullet$$ so the set of edges $G^\star(A)$ of graph $G^\star$ is the set of homomorphisms from graph $G$ to graph $\bullet$. This set is either empty $\emptyset$ --- if $G$ contains any edge, or a singleton $1$ otherwise.

Similarly, the graph associated to the presheaf $\hom(-, B)$ is: $$ \bullet \rightarrow \bullet $$ and so the set of nodes $G^\star(B)$ of graph $G^\star$ is the set of homomorphisms from graph $G$ to graph $\bullet \rightarrow \bullet$. This set is empty iff $G$ contains composable edges (loops $\looparrowright$ or sequences $\bullet \rightarrow \bullet \rightarrow \bullet$). If $G$ does not contain any sequence of composable edges, then the set of nodes is equal to $2^D$, where $D$ is the largest discrete full subgraph of $G$.

If there is an edge in $G^\star(A)$, and there are at least two distinct nodes in $ G^\star(B)$ then the edge is non-degenerated (its source is different from its target).

By duality, the right part ($R$ in Todd's answer) of the Isbell conjugation $(\mathbf{Set}^\mathbb{C})^{op} \to \mathbf{Set}^{\mathbb{C}^{op}}$ is defined on graphs in essentially the same way.

Here are some examples of graphs $G$ with their corresponding graphs $G^\star$: \begin{array}{ccc} \emptyset & \mapsto & \looparrowright \newline \bullet & \mapsto & \bullet \rightarrow \bullet \newline \looparrowright & \mapsto & \emptyset \newline \bullet \rightarrow \bullet & \mapsto & \bullet \newline \bullet \rightarrow \bullet \; \bullet & \mapsto & \bullet \; \bullet \newline \bullet \; \bullet & \mapsto & \bullet \rightarrow \bullet \; \bullet \; \bullet \newline \bullet \rightarrow \bullet \; \bullet \; \bullet & \mapsto & \bullet \; \bullet \; \bullet \; \bullet \newline \end{array}

One may easily verify that, due to size issues (i.e. there are no isomorphism $2^K \approx K$ for any $K$) and the fact that in the image of $(-)^\star$ there are no graphs with more than one edge, only the first four graphs from the above table are fixed-points of the Isbell conjugation. Moreover, these graphs together with their homomorphism constitute a category isomorphic to: $$\overset{0\longrightarrow A}{} \stackrel{\longrightarrow}{\longrightarrow} \overset{B\longrightarrow 1}{}$$

The gap in Todd's proof of (co)completeness is in the implicit assumption that the fixed-points of a monad are the same as fixed-points of the associated idempotent monad. There is an abstract counterexample to this claim --- fixed-points of an idempotent monad constitute a reflective subcategory of the base category, whereas fixed-points of an arbitrary monad generally not (perhaps one of the idea of an associated idempotent monad is just to add "enough" fixed-points to the monad to make the subcategory of fixed-points reflective).