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Short answer: Yes, it can possibly have an adjoint.

Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra. Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.

Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$. It is an algebra morphism. Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor \begin{align} \varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C} \ . \end{align} A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.

Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. preserves coequalizers). Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of $$ r,\ \varepsilon \otimes id_V \colon L \otimes V \to V \ . $$

So for a particular situation where it works: take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO. Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat. Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.

Short answer: Yes, it can possibly have an adjoint.

Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra. Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.

Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$. It is an algebra morphism. Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor \begin{align} \varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C} \ . \end{align} A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.

Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. $L \otimes - $ preserves coequalizers). Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of $$ r,\ \varepsilon \otimes id_V \colon L \otimes V \to V \ . $$

So for a particular situation where it works: take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO. Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat. Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.

Short answer: Yes, it can possibly have an adjoint.

Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra. Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.

Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$. It is an algebra morphism. Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor \begin{align} \varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C} \ . \end{align} A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.

Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. preserves coequalizers). Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of $$ id_1 \otimes r,\ (m_1 \circ 1 \otimes \varepsilon) \otimes id_V \colon 1 \otimes L \otimes V \to 1 \otimes V \ . $$

So for a particular situation where it works: take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO. Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat. Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.

Short answer: Yes, it can possibly have an adjoint.

Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra. Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.

Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$. It is an algebra morphism. Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor \begin{align} \varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C} \ . \end{align} A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.

Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. preserves coequalizers). Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of $$ r,\ \varepsilon \otimes id_V \colon L \otimes V \to V \ . $$

So for a particular situation where it works: take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO. Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat. Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.