Has anyone ever seen a Monad that is very much like the List Monad but is also a co-monad, and also a Frobenius monad? In [this paper](https://arxiv.org/abs/1408.5809) they give examples of List-like monads called Containers and they give one that is a Comonad, namely Trees. The comonad axiom takes each node in the tree and labels it with the tree rooted at that node. In the comments, there is an answer to a previous question of mine where someone suggests a bimonad, so comonad and monad that does not satisfy the frobenius property.



A List-Like Frobenius Monad
I claim the only Frobenius monad on $\mathrm{Set}$ is the trivial monad given by the identity functor (which I guess needless to say isn't "very much like" the List monad). 


Frobenius monads on a category $C$ are essentially the same as monads on $C$ whose underlying endofunctor is left adjoint to itself. For details on this assertion, see the nLab article on [Frobenius monads](https://ncatlab.org/nlab/show/Frobenius+monad). 


Endofunctors on $\mathrm{Set}$ that have right adjoints, i.e., that are left adjoints, are very easily described: they are precisely the endofunctors of the form $X \mapsto S \cdot X$ where $S$ is a set and $S \cdot X$ denotes the coproduct of an $S$-indexed family of copies of $X$; it can be identified with $S \times X$. Natural transformations between such functors are exactly the ones induced by functions $S \to T$. The right adjoint to such a functor is $X \mapsto X^S$. The only case where we have an isomorphism (natural in $X$) $S \cdot X \cong X^S$ is where $S = 1$; this is immediate from the component at $X = 1$. This $S = 1$ corresponds to the identity endofunctor. 


Thus the only endofunctor on $\mathrm{Set}$ adjoint to itself is the identity functor, and this has exactly one monad structure which is the one who unit and multiplication are identity transformations. 


