Monads which preserve equalizers I'm looking for examples of monads that preserve equalizers, and an understanding of sufficient conditions that ensure this.
This is a situation where the advice given at the end of this answer fails to be helpful, since of course any monad on a poset preserves equalizers. Considering the first part of that answer, a related question would be:


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*What conditions on a theory $\mathbb{T}$ ensure that the corresponding monad on $\mathbf{Set}$ preserves equalizers?


This turns out to be quite subtle. For example, the free group and free abelian group monad don't preserve the equalizer diagram,
$$\emptyset \rightarrow \{0,1\}\rightrightarrows \{0,1\},$$
where the parallel morphisms are the two constant maps, since in the latter case the map $\mathbb{Z} \to \mathbb{Z}^2$ sending $n$ to $(n,-n)$ equalizes the corresponding maps $\mathbb{Z}^2 \rightrightarrows \mathbb{Z}^2$ but doesn't factor through the trivial group.
As Max pointed out in the comments, the same is not applicable to monoids, since we have no inverses and so the comparable equalizer in the above is actually preserved. Indeed, for the free monoid monad we have that a word $w$ is in the equalizer of two parallel maps $F(f),F(g):F(X) \rightrightarrows F(Y)$ if and only if each of its constituent letters is (since there is a unique expression for any word), and hence if and only if the word is contained in the free monoid on the equalizer of $f$ and $g$. That is, this monad does preserve equalizers. For the free commutative monoid monad, on the other hand, the equalizer of the identity map and the map $\{0,1\}\to \{0,1\}$ which exchanges the elements is not preserved, since the map $\mathbb{N} \to \mathbb{N}^2$ sending $x$ to $(x,x)$ is the equalizer there!
The covariant powerset monad $P: \mathbf{Set} \to \mathbf{Set}$ doesn't preserve equalizers (as I'd initially thought, thanks Max) since considering the functions $\pi_1,(1-\pi_1):\{0,1\}^2 \rightrightarrows \{0,1\}$ we have $P\pi_1(\{(0,0),(1,1)\}) = \{0,1\} = P(1-\pi_1)(\{(0,0),(1,1)\})$, and similarly for three other possible choices, so the equalizer of these functions is not $P(\emptyset)$. A question which would require more work to check is:


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*Does the double powerset functor (obtained by iterating the contravariant powerset functor) preserve equalizers?



Background: I've been studying the article "Cartesian Monads on Toposes" by Peter Johnstone where he studies monads on toposes preserving all finite limits, and finds that there really aren't very many (on $\mathbf{Set}$, they are in correspondence with strongly zero-dimensional locales, which is to say locales in which every open cover can be refined to a pairwise disjoint one, a very restrictive condition), so one motivation is to understand how much is gained by relaxing the conditions on the monad. 
 A: Turning my earlier comment into an answer, every Cartesian monad preserves equalizers, in the sense of a monad on a Cartesian category whose functor part preserves pullbacks and whose naturality squares induced by the multiplication and unit are all pullbacks. 
There are several examples of Cartesian monads on the nLab page, and there is a beautiful characterization due to Carboni and Johnstone of the Cartesian monads arising from algebraic theories: the appropriate theories are the strongly regular ones, those which can be axiomatized using equations in which the same variables appear in the same order on both sides. So associativity is OK, but commutativity and inverses are not, recovering precisely the examples discussed by you and Max.  
The reason every Cartesian monad preserves equalizers is simply that any pullback-preserving functor $F:A\to B$ whose domain admits a terminal object preserves all finite connected limits, including equalizers. The argument is that the canonical lift of $F$ to a functor $\hat F:A\to B\downarrow F(1)$ preserves the terminal object (by construction) and pullbacks (because $F$ does), so preserves equalizers; but the forgetful functor $B\downarrow F(1)\to B$ preserves equalizers, so $F$ does as well. 
