Free idempotent monad associated to a monad Let $C$ be a category. There is a full subcategory $\text{IdemMnd}(C) \hookrightarrow \text{Mnd}(C)$ of the category of monads on $C$ spanned by the idempotent monads. Given a monad $T$ on $C$, suppose we wish to construct an idempotent monad $\hat T$ on $C$ associated to $T$. There are two ways by which we might consider doing so.

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*Considering the value at $T$ of a left adjoint to the inclusion $\text{IdemMnd}(C) \hookrightarrow \text{Mnd}(C)$ (the free idempotent monad).

*Considering the value at $T$ of a right adjoint to the inclusion $\text{IdemMnd}(C) \hookrightarrow \text{Mnd}(C)$ (the cofree idempotent monad).

The second case is well-studied: for instance, it is studied by Fakir in Monade idempotente associée à une monade.
Has the first case been studied in the literature? (Alternatively: is there a good reason it has not?) I spell out the universal property we expect of $\hat T$ below, to make it easier to compare with reformulations that may appear.

Let $T$ be a monad on $C$. The free idempotent monad on $T$, denoted $\hat T$ when it exists, is an idempotent monad on $C$ equipped with a monad morphism $\tau : T \to \hat T$ such that for every idempotent monad $T'$ on $C$ and monad morphism $\phi : T \to T'$, there exists a unique monad morphism $\hat\phi : \hat T \to T'$ such that $\hat\phi \circ \tau = \phi$.
We may rewrite this in terms of the categories of algebras as follows: for every functor $F : T'\text{-Alg} \to T\text{-Alg}$ commuting with the forgetful functors, there exists a unique functor $\hat F : T'\text{-Alg} \to \hat T\text{-Alg}$ commuting with the forgetful functors such that $\tau\text{-Alg} \circ \hat F = F$.
 A: What follows is a long comment, and because the question is asking for a reference, it is also an irrelevant comment.

Anyhow, the whole idea behind the Fakir construction is to build on a characterisation of idempotent monads.
Prop.
A monad $T$ is idempotent iff $T\eta, \eta T: T \to T^2$ are the same arrow.
Following this characterisation, Fakir thought that an intelligent way to make a monad idempotent was to take the equalizer $T_1$ of $T\eta$ and $\eta T$,$$ T_1 \hookrightarrow T \rightrightarrows T^2. $$ Now $T_1$ is a monad, but it might not be idempotent (yet), so we have to transfinitely iterate this procedure and hope that it converges at some point.
So, if I would have to guess a dual construction, I would take the coequalizer,
$$T \rightrightarrows T^2 \twoheadrightarrow T_1.$$
The key thing to check is that $T_1$ still produces a monad. It is clear that we have a unit $1 \to T \to T_1$. Concerning the multiplication, that is a bit more delicate and one should dualize the argument discussed here. I could believe it works, but a gut feeling prevents me from true optimism.
