I was wondering if there's any known result which classifies which functors $F : \mathcal{B} \to \mathcal{C}$, where $\mathcal{B}$ and $\mathcal{C}$ are both varieties of algebras, are monadic.
I know that we have this result (though I unfortunately forgot what's it's called): $F : \mathcal{B} \to \mathcal{C}$ has a left adjoint if and only if it is representable by a co-$\mathcal{C}$ object of $\mathcal{B}$. So, it would be equivalent to ask for which such objects $X$ the functor $\operatorname{Hom}_{\mathcal{B}}(X, -) : \mathcal{B} \to \mathcal{C}$ is monadic.
I do know that it's not automatically monadic. For example, the group of units functor $\mathbf{Rings} \to \mathbf{Groups}$ is not monadic since it does not reflect isomorphisms (e.g. it sends the non-isomorphism $\mathbb{C} \hookrightarrow \mathbb{C}[t]$ to an isomorphism). On the other hand, it does have a left adjoint $G \mapsto \mathbb{Z}[G]$. (And it is represented by $X = \mathbb{Z}[t, t^{-1}]$, with cogroup structure given by coidentity $X \to \mathbb{Z}$, $t \mapsto 1, t^{-1} \mapsto 1$; coinverse $X \to X$, $t \mapsto t^{-1}, t^{-1} \mapsto t$; and comultiplication $X \to X \amalg X \simeq \mathbb{Z} \langle t, t^{-1}, u, u^{-1} \rangle$, $t \mapsto tu$, $t^{-1} \mapsto u^{-1} t^{-1}$.)
On the other hand, there are plenty of examples where it is monadic. For example, for any variety of algebras $\mathcal{C}$, the underlying set functor $\mathcal{C} \to \mathbf{Sets}$ is monadic. In fact, I think any "forgetful" functor which just takes a subset of the operations and relations from the domain should be monadic. (Though this last isn't a categorical condition - for example, the functor $\mathbf{Groups} \to \mathbf{BinOp}$ which takes $G = (|G|, \cdot, ^{-1}, e) \mapsto (|G|, (x, y) \mapsto x y^{-1})$ should also be monadic, since $\mathbf{Groups}$ is equivalent to the variety with a binary operation $/$, a constant $e$, and the relations $x/x = e$, $x/e = x$, $(x/z) / (y/z) = x/y$.)
Also, if I'm not mistaken, the functor $\mathbf{Sets} \to \mathbf{Sets}$, $X \mapsto X \times X$, is monadic, giving an example which doesn't respect "underlying sets".