Is power set functor determined by its image on objects? Let $\mathbf{Set}$ be a category of set, and $\mathcal{P}:\mathbf{Set} \to \mathbf{Set}$ be a power set functor defined as
$$
\mathcal{P}(X) = 2^{X} = \{U\subseteq X\}
$$
and for any function $f:X\to Y$, 
$$
\mathcal{P}(f): \mathcal{P}(X) \to \mathcal{P}(Y), \quad \mathcal{P}(f)(U) = f(U). 
$$
It is not hard to show that this really defines a functor. My question is:

Let $\mathcal{F}:\mathbf{Set}\to \mathbf{Set}$ be a functor that satisfies $\mathcal{F}(X) = \mathcal{P}(X)$ for all set $X$. Does this imply that $\mathcal{F} = \mathcal{P}$? i.e. $\mathcal{F}(f) =\mathcal{P}(f)$ for all morphisms (functions) $f$?

I believe that this is true, but I have no idea about proof. This question is motivated from the functional programming, especially about list functor (in Haskell). This is the question that I posted on Haskell reddit. My strategy is that, if one can show that the above statement is true, then we can modify the proof to give an answer to the original question about list functor. Maybe one can try to show for the category of finite sets or countable sets. 
 A: Here's a class of counter-examples:
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
$$
\mathcal F(f) = r_Y\circ \mathcal P(f) \circ r_X^{-1}.
$$
You can check that this is a functor and one non-trivial choice of $r_X$ would be taking complements, i.e. $r_X(U)=X\setminus U$, then $\mathcal Ff(U) = Y\setminus f(X\setminus U)$.
A: There exists at least one other endofunctor of $\mathbf{Set}$ that sends every set to its powerset. This endofunctor sends a function $f:X\to Y$ to
$$\widehat{f} :P(X)\to P(Y):U\mapsto \widehat{f}(U)=\{y\in Y\mid f^{-1}(\{y\})\subset U\}$$
(where $f^{-1}$ is the inverse image).
One can check directly that $\widehat{f\circ g}=\widehat{f}\circ \widehat{g}$ and $\widehat{id_X}=id_{P(X)}$, or use the following fact (which explains the origin of that definition) : for every set $X$, the powerset $P(x)$ is a poset (ordered by inclusion), and for any given $f$, $P(f), f^{-1}$ and $\widehat{f}$ are all monotone functions and we have two adjunctions $P(f)\dashv f^{-1}\dashv \widehat{f}$. Then, for any $g$ we have a chain of adjunctions
$$P(f\circ g)\dashv (f\circ g)^{-1}\dashv \widehat{f\circ g}$$
and since adjunctions can be composed, we also have
$$P(f)\circ P( g)\dashv g^{-1} \circ f^{-1}\dashv \widehat{f}\circ \widehat{g}$$
Since $P$ is a functor, the first term of the two chains coincide. By uniqueness of adjoint functors the other terms also coincide, thus $\widehat{f\circ g}=\widehat{f}\circ \widehat{g}$. You can use a similar argument for the identities.
