Is power set functor determined by its image on objects?

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.

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 $\mathcal Ff(U) = Y\setminus f(X\setminus U)$.