When does transporting a monad along an adjunction "preserve" its category of modules? If I have an adjunction $F \dashv G$ where $F \colon C \to D$, and $(N, m, u)$ is a monad on $D$, then I can define a monad $\widetilde{N}$ on $C$ via
\begin{align}
   (GNF, \widetilde{m} = G m_F \circ GN \varepsilon_{NF}, \widetilde{u} = G u_F \circ \eta)
\ ,
\end{align}
where $\eta$ and $\varepsilon$ are the unit and counit of the adjunction.
My question is:

*

*When is the Eilenberg-Moore category $D^N$ of $N$ equivalent to $C^{\tilde{N}}$?

Maybe that's a bit too hard, a possibly simpler question is:


*There is an obvious functor $D^N \to C^{\widetilde{N}}$, sending and $N$-module $(d, \rho \colon Nd \to d)$ to the $\widetilde{N}$-module $(Gd, G\rho \circ GN\varepsilon_d)$.
Can we say when this functor is an equivalence?

(PS: please don't say "Take $C = D$ and $F = G = id_C$." haha)
 A: A very satisfying partial answer along the lines of what Arnaud suggested is given in Hopf monads on monoidal categories (Bruguieres, Lack, Virelizier).
Namely, let $T$ be a monad on a category $C$, and let $N$ be an endofunctor on $C^T$.
Then the construction I performed above uses the free-forgetful adjunction of $F_T \dashv U_T \colon C \to C^T$, and yields what they call the pushforward $(F_T, U_T)_* N := N T$, which is an endofunctor on $C$.
Now, if $N$ is a monad, the pushforward will (as I showed above) be a monad on $C$, and it is called the cross product $N \rtimes T$.
In the paper it is then said that a sufficient condition for the comparison functor
\begin{align}
K_{N \rtimes T} \colon (C^T)^N \to C^{N \rtimes T}
\end{align}
to be an isomorphism is that $N$ preserve reflexive coequalizers.
And as a side note, this cross product construction really is the usual one:
consider a bialgebra $B$ over a commutative ring $k$, and let $A$ be a $B$-module algebra.
There is the cross or smash product construction, yielding a new $k$-algebra $A \rtimes B$ such that $A-(B-\text{mod}) = (A \rtimes B)-mod$.
Now, algebras are monads: $B \otimes - $ is a monad on $k$-modules - its modules are exactly the $B$-modules -, and $A \otimes -$ is a monad on $B$-modules.
Then
\begin{align}
   (A \rtimes B) \otimes - = (A \otimes -) \rtimes (B \rtimes -)
\end{align}
as monads, and of course
\begin{align}
(k\text{-mod}^{B \otimes -})^{A \otimes -} =
k\text{-mod}^{(A \rtimes B )\otimes -}
\end{align}
I find this immensely satisfying.
