For shorthand, denote: $\mathcal{P}^{*}(A) = \mathcal{P}(A) \setminus \{\emptyset\}$ .
Additionally, use $F_{f}(x)$ as shorthand for $F(f)(x)$.
We have a functor: $$F:(A\to B) \to (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B)) $$
defined by: $$\forall f:A\to B,\;\; F_{f}(X) = \{ f(x) \mid x\in X \} $$
This "lifts" bijections between $A$ and $B$ to bijections between $\mathcal{P}^{*}(A)$ and $\mathcal{P}^{*}(B)$ in a "natural" fashion.
We know it is a functor because given $f:A\to B$ and $g: B\to C$, the following $2$ properties hold:
$$ F_{g}\circ F_{f} = F_{g \;\circ \;f}\\
F_{(\text{id}_{X})} = \text{id}_{(\mathcal{P}^{*}(X))}$$
where $\text{id}_{X} : X \to X$ denotes the identity function on a set $X$.
Compositionality makes $F$ a functor.
We would like to find a functor $G$ that "undos" $F$, in a sense. That is:
$$G: (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B)) \to (A\to B)
\\G_{F_{f}} = f$$
where $G$ is a functor, meaning it obeys:
$$
G_{g}\circ G_{f} = G_{g \;\circ \;f}\\
G_{\text{id}_{(\mathcal{P}^{*}(X))}} = \text{id}_{X}$$
This hypothetical functor would have all the "nice" properties we expect from a way to bring bijections of $ (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B))$ down to bijections of $A \to B$, including the "relabeling property":
If you start with a bijection from $(\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B))$, permute the labels of the elements of $A$ (or the labels of the elements of $B$) in it amongst themselves, then bring it "down" to a bijection from $A \to B$ it should get the same result as bringing it down first, then applying the specified relabelings.
Formally, given $a: A \to A$, $\;b: B \to B$, $\;f:\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B)$, we have:
$$G_{F_{b} \; \circ \; f \; \circ \; F_{a}} = b \; \circ \; G_{f} \; \circ \; a $$
This follows from the previously stated properties of $G$.
Now that we've formalized what we're looking for, let's prove it does not exist.
Theorem 1
Given $2$ bijections $x: A\to B$, $y : A\to B$, the preimage of $G$ with respect to $x$ has the same cardinality as the preimage of $G$ with respect to $y$. That is, for $I_{x} = \{X \mid G_X = x\}$, $I_{y}= \{Y \mid G_{Y} = y\}$, we have $|I_{x}| = |I_{y}|$.
Proof:
Let: $\psi : B \to B$ have the property of $\psi \circ x = y$. Then $F_{\psi}$ can give us a bijection (through composition), between $I_{x}$ and $I_{y}$. Thus $|I_{x}| = |I_{y}|$. Any two bijections from $A$ to $B$ have an equal number of elements (the elements being specific bijections from $(\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B))$) in their associated preimages with respect to $G$.
Theorem 2
Let there be a set $A$ such that $|A| \geq 3$.
Then a functor $G: (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(A)) \to (A\to A)$ cannot exist.
Notes: For $|A| = 2$ you can construct $G$. There is a "nice" mapping $G: (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(B)) \to (A\to B)$ when $|A| = |B| = 2$. I will first give the impossibility proof for $|A| = 3$, then give a general argument.
Proof:
There are $3! = 6$ bijections from $A$ to $A$.
There are $(2^3-1)! = 5040$ bijections from $(\mathcal{P}^{*}(A)$ to $\mathcal{P}^{*}(A))$.
The inverse image of any $A \to A$ bijection with respect to $G$ is of cardinality $5040/ 6 = 840$.
We will be studying $|I_{\text{id}_{A}}|$, the cardinality of the set of $(\mathcal{P}^{*}(A)$ to $\mathcal{P}^{*}(A))$ bijections which $G$ maps to $\text{id}_{A}$.
Denote: $ f^n = \underbrace{f\;\circ\;f\;\circ \ldots f}_{n\ \text{times}}$.
The bijections of $f:(\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(A))$ can be considered as permutations of $7$ elements.
Consider those which cycle $5$ of the elements and leave the rest unchanged.
Denote one of these as $f_5$.
We have ${f_{5}}^5 = \text{id}_{\mathcal{P}^{*}(A)}$.
So ${G_{f_{5}}}^5 = G_{{f_{5}}^5} = G_{{\text{id}_{{P}^{*}(A)}}} = \text{id}_{A}$ .
But $G_{f_{5}}$ is only a bijection from $A$ to $A$, a $3$ element set.
It cannot have a nontrivial cycle of a length which divides $5$. Therefore $G_{f_{5}} = \text{id}_{A}$.
We can make a similar argument for $G_{f_{7}}$, where $f_{7}:(\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(A))$ denotes a given $7$-cycle of $\mathcal{P}^{*}(A)$. We have $G_{f_{7}} = \text{id}_{A}$.
The number of $5$-cycles of $\mathcal{P}^{*}(A)$ is $\binom{7}{5}*4! = 504$.
The number of $7$-cycles of $\mathcal{P}^{*}(A)$ is $6! = 720$.
From this we get $|I_{\text{id}_{A}}| \geq 504 + 720 > 840$, contradicting theorem $1$.
We cannot construct the functor $G: (\mathcal{P}^{*}(A) \to \mathcal{P}^{*}(A)) \to (A\to A)$ when $|A| = 3$.
Now for the general case. We will rely on $2$ lemmas:
Lemma 1
For $n > 2$, there exists a prime $p$ such that $n < p < 2^n$.
This follows from the Bertrand–Chebyshev theorem.
Lemma 2
Let $n$ be a natural number, and $i > 1$ be an odd number. Then the $i$-cycles $f_{i}$ (on a set of $n$ elements) generate the alternating group $\mathcal{A}_{n}$.
Proof:
Let $f_{i}$ be a $i$-cycle and $\alpha_{k ,f_i(k)}$ be an inversion of $2$ consecutive elements of the cycle $f_{i}$. Then $g_{i} = \alpha_{k ,f_i(k)} \; \circ \; f_{i} \circ \; \alpha_{k ,f_i(k)}$ will also be an $i$-cycle, and $f_{i}\circ g^{-1}_{i}$ will be a $3$-cycle $(f^{-1}(k), \; k, \; f(k))$. Thus we can generate all $3$-cycles, which generate the alternating group $\mathcal{A}_{n}$, which contains $f_{i}$, since odd cycles have even parity.
Now for the proof:
Let $A$ be a set with $|A| = n >3$.
Due to theorem $1$, we have $|I_{\text{id}_{A}}| = \frac{(2^n-1)!}{n!}$.
Due to lemma $1$, there exists a prime $p$ with $n < p < 2^n$. Generalizing the composition argument of the previous proof, we know that $G_{f_{p}} = \text{id}_{A}$ for all $p$-cycles $f_{p}$ of $\mathcal{P}^{*}(A)$.
Due to lemma 2 and the composition property of $G$, we can say: $\forall \alpha\in \mathcal{A}_{(\mathcal{P}^{*}(A))},\; G_{\alpha} = \text{id}_{A}$.
Ie. $G$ will map any permutation which is of the alternating group of $\mathcal{P}^{*}(A)$ to $\text{id}_{A}$.
We have $|I_{\text{id}_A}| \geq |\mathcal{A}_{(\mathcal{P}^{*}(A))}| = \frac{(2^n-1)!}{2} > \frac{(2^n-1)!}{n!}$ for $n > 2$, contradicting theorem $1$. This completes the proof.
