Recall the powerset functor $\mathcal{P}:\mathbf{Set} \to \mathbf{Set}$ defined as
- $\mathcal{P}(X) = 2^{X} = \{U\subseteq X\}$ on objects
- $\mathcal{P}(f: X \to Y): \mathcal{P}(X) \to \mathcal{P}(Y),\ \ \ \mathcal{P}(f)(U) = f(U)$ on morphisms
Are there functors $\mathcal{F}:\mathbf{Set}\to \mathbf{Set}$ agreeing with $\mathcal{P}$ on objects, i.e. $\mathcal{F}(X) = \mathcal{P}(X)$ for all sets $X$?
This question was answered in "Is power set functor determined by its image on objects?".
However, all functors given there are naturally isomorphic to $\mathcal{P}$.
Question: Are there any such functors which are not naturally isomorphic to $\mathcal{P}$?
Summary of Results
All functors given in the question linked above can be succinctly characterized as follows:
For each set $X$ choose a bijection $r_X\colon \mathcal P(X)\to\mathcal P(X)$. Now let your functor $\mathcal F$ be defined on morphisms $f\colon X\to Y$ by $$\mathcal F(f) = r_Y\circ \mathcal P(f) \circ r_X^{-1}$$
One non-trivial choice of $r_X$ would be taking complements, i.e. $r_X(U)=X\setminus U$, then $\mathcal Ff(U) = Y\setminus f(X\setminus U)$.
As per my comment on Christoph's post, we can exhibit the natural isomorphism to $\mathcal{P}$ as follows:
$$ \require{AMScd} \begin{CD} \mathcal{P}(X) @>{r_X}>> \mathcal{P}(X)\\ @V{\mathcal{P}f}VV @VV{\mathcal{F}f = r_y\ \circ\ \mathcal{P}f\ \circ\ r_X^{-1}}V \\ \mathcal{P}(Y) @>{r_Y}>> \mathcal{P}(Y)\end{CD} $$
