If I understood the OP correctly, the problem can be stated as follows :

Problem 1. Let $X$ be a set and $f$ be any map ${\cal P}(X) \to X$ , and $A$ be defined as above ( $A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace$ ). Find a definable $B$ (in termes of $F$ and $X$) such that $B \neq A$ and $F(B)=F(A)$.

Now Problem 1 is equivalent to the simpler problem :

Problem 2. Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find a definable $y_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.

The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w_k \rbrace_{k \geq 1}$ outside of $Y$. Now define $X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define $F : {\cal P}(X) \to X$ by
 

1. $F(\lbrace (y,w_1) \rbrace)=a$, if $y\in {Y'} $
    
2. $F(\lbrace (y,w_{k}) \rbrace)=(y,w_{k-1})$ , for all $y\in Y$ and $k\geq 2$
    
3. $F(X)=a$
    
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$)
    .

Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w_1) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.

If I understood the OP correctly, the problem can be stated as follows :

Problem 1. Let $X$ be a set, let $F:{\cal P}(X) \to X$, and let $A$ be defined as above: $$A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace.$$ Find a definable $B$ (in terms of $F$) such that $B \neq A$ and $F(B)=F(A)$.

Now Problem 1 is equivalent to the simpler problem :

Problem 2. Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find a definable $y_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.

The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w_k \rbrace_{k \geq 0}$ outside of $Y$. Now define $X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define $F : {\cal P}(X) \to X$ by:

1. $F(\lbrace (y,w_0) \rbrace)=a$, if $y\in {Y'} $,
2. $F(\lbrace (y,w_{k+1}) \rbrace)=(y,w_k)$ , for all $y\in Y$ and $k\ge0$,
3. $F(X)=a$, and
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$).

Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w_0) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.

If I understood the OP correctly, the problem can be stated as follows :

Problem 1. Let $X$ be a set and $f$ be any map ${\cal P}(X) \to X$ , and $A$ be defined as above ( $A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace$ ). Find a definable $B$ (in termes of $F$ and $X$) such that $B \neq A$ and $F(B)=F(A)$.

Now Problem 1 is equivalent to the simpler problem :

Problem 2. Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find a definable $y_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.

The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w_k \rbrace_{k \geq 1}$ outside of $Y$. Now define $X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define $F : {\cal P}(X) \to X$ by

1. $F(\lbrace (y,w_1) \rbrace)=a$, if $y\in Y'$
   
2. $F(\lbrace (y,w_{k}) \rbrace)=(y,w_{k-1})$ , for all $y\in Y$ and $k\geq 2$
   
3. $F(X)=a$
   
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$)
   .

Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w_1) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.

If I understood the OP correctly, the problem can be stated as follows :

Problem 1. Let $X$ be a set and $f$ be any map ${\cal P}(X) \to X$ , and $A$ be defined as above ( $A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace$ ). Find a definable $B$ (in termes of $F$ and $X$) such that $B \neq A$ and $F(B)=F(A)$.

Now Problem 1 is equivalent to the simpler problem :

Problem 2. Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find a definable $y_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.

The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w_k \rbrace_{k \geq 1}$ outside of $Y$. Now define $X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define $F : {\cal P}(X) \to X$ by

1. $F(\lbrace (y,w_1) \rbrace)=a$, if $y\in {Y'} $
   
2. $F(\lbrace (y,w_{k}) \rbrace)=(y,w_{k-1})$ , for all $y\in Y$ and $k\geq 2$
   
3. $F(X)=a$
   
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$)
   .

Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w_1) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.