It is well-known that the composite of monadic functors $U: C \to C'$ and $U': C' \to C''$ need not be monadic. One standard example is the forgetful functor $\mathrm{Cat} \to \mathrm{RefGph}$ from categories to reflexive graphs, followed by the functor $\mathrm{RefGph} \to \mathrm{Set}$ that takes a reflexive graph to its set of arrows.
Another example is by taking the full inclusion of torsionfree abelian groups in abelian groups, followed by the forgetful functor from abelian groups to sets. The first is monadic because it is a reflective subcategory inclusion. But the composite, the forgetful functor from torsionfree abelian groups to sets, is not monadic. Another example in this vein: the forgetful functor from commutative rings with no nonzero nilpotents to sets is not monadic.
For every example of this phenomenon I've seen so far, there is a uniform and very easy reason for why $U'U$ is not monadic. Let $F$ and $F'$ be the left adjoints to $U: C \to C'$ and $U': C' \to C''$, respectively. Then $U'F'$ and $(U'U)(FF')$ are both monads on $C''$, and there is a morphism of monads
$$U'\eta F': U'F' \to U'UFF'$$
where $\eta$ denotes the unit of $F \dashv U$. In every case I've seen, it happens that this monad morphism is an isomorphism. Consequence: if $U'U$ were in fact monadic, then $C$ and $C'$ would both be equivalent to the Eilenberg-Moore category of this same (up to iso) monad on $C''$, i.e., the comparison functor $U: C \to C'$ would have to be an equivalence. But it's manifestly not, for each of these examples.
Call this, that $U'\eta F'$ is an isomorphism but $U$ is not an equivalence, a "trivial" reason for why a composite of monadic functors need not be monadic. Can anyone give me a nontrivial example, where the composite $U'U$ isn't monadic but where $U'\eta F'$ is not an isomorphism?
