"Equivalence" is to "group" as "adjoint" is to ....?

If everything has both a left and a right adjoint, then you're talking about a rigid monoidal category. (Here I've done the usual dimension shift where a 2-category with one object is the same as a monoidal category.)

If you only want adjoints on one side it's a bit more awkward because that's not left rigid (since the left adjoint will only have a right adjoint) nor right rigid. But for the single functor case it'd be a monoidal category generated by a right rigid object.

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.