Do free monads on $Set^I$ (functor category) always exist? To be precise I'm asking if for every $I$ there's a left adjoint to the forgetful functor $U : Mnd(Set^I) \to End(Set^I)$ where $Mnd(C)$ is the category of monads on $C$ and $End(C)$ the one of endofunctors.
http://ncatlab.org/nlab/show/free+monad lists some conditions for the existence of free monads but I can't immediately tell if they apply here.
 A: They exist in various special cases.
For any small category $\mathcal{C}$, the functor category $[\mathcal{C}, \textbf{Set}]$ is a locally finitely presentable category, so for any finitary endofunctor $J : [\mathcal{C}, \textbf{Set}] \to [\mathcal{C}, \textbf{Set}]$ (i.e. an endofunctor preserving filtered colimits), the category of $J$-algebras is strictly monadic over $[\mathcal{C}, \textbf{Set}]$, and therefore by Barr's theorem, the free monad on $J$ exists.
Since any locally finitely presentable category is also locally $\kappa$-presentable for all regular infinite cardinals $\kappa$, one can further show that any accessible endofunctor on $[\mathcal{C}, \textbf{Set}]$ has a free monad.
Nonetheless, not every endofunctor is accessible. For example, the covariant powerset functor $\mathscr{P} : \textbf{Set} \to \textbf{Set}$ is not accessible. (This is basically Cantor's theorem.) Moreover, the category of $\mathscr{P}$-algebras does not have an initial object: if $X$ is any non-empty $\mathscr{P}$-algebra, then we can always find a $\mathscr{P}$-algebra $Y$ such that there are at least two $\mathscr{P}$-algebra homomorphisms $X \to Y$. In particular, the category of $\mathscr{P}$-algebras cannot be strictly monadic over $\textbf{Set}$, since any category monadic over $\textbf{Set}$ has an initial object. (This is Example 6.8 in [Barr, 1970, Coequalisers and free triples].) However, since $\textbf{Set}$ is complete, the converse of Barr's theorem applies (see Proposition 22.4 in [Kelly, 1980, A unified treatment of transfinite constructions]), so the free monad on $\mathscr{P}$ cannot exist.
