Existence of non-trivial idempotent endofunctor on $\mathcal{Set}$ Endofunctors on the category of sets seem surprisingly hard to study (see for example this other question about the existence of a functor sending any cardinal on the next cardinal.) Here's another question that I wasn't able to answer:
Is there a functor $F : \mathcal{Set} \to \mathcal{Set}$ such that $ F \circ F = F$, but which is not the identity and which is not constant when restricted to the non-empty sets? Edit: Moreover, I would want a functor which is not naturally isomorphic to the identity or to a functor constant on the non-empty sets.
If we only ask that it is not constant, for any non-empty set $S$, we can define a functor as $F(\emptyset) = \emptyset$ and $F(A) = S$ for $A$ non-empty, and for any function $f: A \to B$ between non-empty sets, $F(f) = \mathrm{id}_S$.
Such a functor, if it exists, must preserve injections and surjections, because it preserves left and right inverses, hence it preserves the order of the non-zero cardinals (but perhaps not strictly).
Assuming axiom of choice for proper classes (as usual in category theory), the category of sets is equivalent to the full subcategory of cardinals, hence it is sufficient to define the functor for this subcategory.
Edit: if we only ask $F$ naturally isomorphic to $F \circ F$, it is also easy to find examples. There are some in Zhen Lin's answer, but another one is $- \coprod A$, with $ A $ any infinite set.
 A: Choose an infinite set $I$ and a bijection $h : I \times I \cong I$.
Then the endofunctors $(-) \times I$ and $(-)^I$ are idempotent up to isomorphism: there are natural isomorphisms $(X \times I) \times I \cong X \times (I \times I)$ and $(X^I)^I \cong X^{I \times I}$, and then the chosen bijection gives "natural" isomorphisms $X \times h : X \times (I \times I) \cong X \times I$ and $X^{h^{-1}} : X^{I \times I} \cong X^I$.
(This is natural in $X$, which is what matters here; non-naturality in $I$ is not relevant.)
It is unclear to me whether an endofunctor $F$ that is idempotent up to isomorphism can always be "improved" to a strictly idempotent functor.
Suppose there is a strictly idempotent functor $G$ and an isomorphism $\theta : F \cong G$.
Then the composite
$$F F X \xrightarrow{F \theta_X} F G X \xrightarrow{\theta_{G X}} G G X = G X \xrightarrow{\theta^{-1}_X} F X$$
is an isomorphism $\phi : F F X \cong F X$ natural in $X$.
This isomorphism automatically satisfies the equation $F \phi_X = \phi_{F X}$, but this is not automatic for a general isomorphism $F F \cong F$.
Indeed, in the context of the example above, $I \times h = h \times I$ means $h (a, b) = a$ and $h (b, c) = c$, which contradicts the assumption that $I$ is infinite.
Yet, this is not quite enough to prove that $(-) \times I$ is not isomorphic to a strictly idempotent endofunctor – it just means that our chosen natural isomorphism $(X \times I) \times I \cong X \times I$ cannot arise by transport of structure from a strictly idempotent endofunctor; perhaps there is some other natural isomorphism that does satisfy the equation.
