Let $\mathrm{Cob}_2$ be the symmetric monoidal $(\infty,1)$-category whose objects are closed oriented $1$-manifolds and whose morphisms are compact oriented surface bordisms. (Higher morphisms are diffeomorphisms, isotopies, etc.)

Another question asked whether symmetric monoidal $(\infty,1)$-functors $\mathrm{Cob}_2 \to \mathcal{C}$ have been classified for general $\mathcal{C}$. The answer seems to be that we have some idea of what's going on, but we don't have a full classification. I'm now wondering whether we know enough to say anything at all, when the target category is $\mathrm{Cob}_2$ itself:

What is the $(\infty,1)$-category of (strongly) symmetric monoidal endo-functors of this $(\infty,1)$-category: $$ \mathrm{Fun}_{(\infty,1)}^\otimes(\mathrm{Cob}_2, \mathrm{Cob}_2) = \ ? $$

I'm asking this because $\mathrm{Cob}_2$ is the simplest bordism category that is not covered by the cobordism hypothesis. (For the extended surface category $\mathrm{Cob}_{\langle 0,1,2\rangle}$ (an $(\infty,2)$-category) this question could be answered by invoking the cobordism hypothesis.) So this seems like a good way of testing how well we understand $\infty$-categories of bordisms beyond the omnipresent cobordism hypothesis.

There are some things one can show about this category of functors by elementary means. For example, because $\mathrm{Cob}_2$ has duals, every natural transformations between symmetric monoidal functors is invertible. Hence the functor category is actually an $\infty$-groupoid and one could equivalently ask: what is its homotopy type?

One can also construct a few examples. Let $A$ be compact oriented $0$-manifold (= a finite set $A$ with a map $A \to \{\pm 1\}$). Then we can construct a functor: $$ \mathcal{Z}_A: \mathrm{Cob}_2 \longrightarrow \mathrm{Cob}_2, \quad M \mapsto M \times A, \quad (W:M \to N) \mapsto (W \times A:M \times A \to N \times A). $$ This defines a functor from the groupoid of compact oriented $0$-manifolds to the functor category I'm interested in. One can also almost define a splitting for this, by sending an arbitrary $\mathcal{Z}:\mathrm{Cob}_2 \to \mathrm{Cob}_2$ to $\pi_0(\mathcal{Z}(S^1))$, though this does not recover the orientation. (Fun fact: when looking at the homotopy category the functor $h\mathrm{Cob}_2 \to h\mathrm{Cob}_2$ that reverses the orientation is isomorphic to the identity functor via a unique natural isomorphism.)

I would suspect that every endofunctor is of the form $\mathcal{Z}_A$ and that we have an equivalence: $$ \mathrm{Fin}_{/\{\pm1\}}^{\cong} \xrightarrow{\ \simeq\ } \mathrm{Fun}_{(\infty,1)}^\otimes(\mathrm{Cob}_2, \mathrm{Cob}_2), \qquad A \mapsto \mathcal{Z}_A. $$ However, there seem to be no tools available to actually prove this.