TITLE: "Equivalence" is to "group" as "adjoint" is to ....?
QUESTION [8 upvotes]: The collection of all self-equivalences of a category $C$ constitutes a $2$-group, which is a categorification of the notion of a group. My question is about what happens when one replaces equivalences by general adjoint functors. 
The n-lab page on adjunctions says:

a morphism in an adjunction need not be invertible, but it has in some sense a left inverse from below and a right inverse from above.

I would like to know if this can be formalised in an interesting way.
In order to ask a specific question, I want to take a single adjoint pair of functors $(L,R):C\to C$, and let $M$ be the monoid of endofunctors of $C$ generated by $L$ and $R$ (together with some suitably chosen family of natural transformations).

Q. What kind of ($2$-)algebraic structure is $M$?

I apologise that this is a somewhat vague question. An ideal answer would have the form "$M$ is a (lax/colax/...) $2$-blah", where "blah" is an interesting algebraic structure that arises in other contexts not a priori having anything to do with adjoint functors. If it makes more sense to change the setting a little---e.g., to consider instead of $M$ the collection of all functors on $C$ admitting a two-sided adjoint---then please go ahead and do so.

REPLY [3 votes]: The free adjunction, i.e. the 2-category generated by an adjunction $C^\to_\leftarrow D$ (with $C$ and $D$ distinct) is described by Schanuel and Street. In particular, if you restrict to the full sub-2-category on $C$, you get the free monad, and the free full sub-2-category on $D$ is the free comonad. The free monad itself is an interesting 2-category; it's the delooping of the strict mononoidal category $\Delta_a$ of augmented simplices, otherwise known as finite linear orders. Since a monad is just a monoid in the endomorphism category, this is equivalent to the fact that $\Delta_a$ is the free strict monoidal category on a monoid (see the last link).
The free adjunction with $C=D$ will be more complicated. I suspect it's not that much more complicated, but I don't know of a reference where it's written down.