Let $S = \{1, 2, ..., 6\}$. We want to find the number of functions $F$ mapping subsets of $S$ to subsets of $S$ such that for any subsets $A, B \subseteq S$, $F(F(A) \cup B) = A \cap F(B)$.

1. Problem Description

Let $S = \{1, 2, ..., 6\}$. We want to find the number of functions $F$ mapping subsets of $S$ to subsets of $S$ such that for any subsets $A, B \subseteq S$, $F(F(A) \cup B) = A \cap F(B)$.

2. Solution Steps

Let $A = \emptyset$. Then $F(F(\emptyset) \cup B) = \emptyset \cap F(B) = \emptyset$ for any $B \subseteq S$.

Let $C = F(\emptyset)$. Then $F(C \cup B) = \emptyset$ for any $B \subseteq S$.

Let $B = \emptyset$. Then $F(C) = \emptyset$.

Let $B = S$. Then $F(C \cup S) = F(S) = \emptyset$.

Let $B = \emptyset$. Then $F(F(A)) = A \cap F(\emptyset) = A \cap C$.

Applying $F$ again, we get $F(F(F(A))) = F(A \cap C)$.

On the other hand, $F(F(F(A))) = F(F(F(A) \cup \emptyset)) = F(A) \cap F(\emptyset) = F(A) \cap C$.

Thus, $F(A \cap C) = F(A) \cap C$.

We have $F(F(A) \cup B) = A \cap F(B)$. Let $A = S$. Then $F(F(S) \cup B) = S \cap F(B) = F(B)$. Since $F(S) = \emptyset$, $F(\emptyset \cup B) = F(B)$. Thus $F(B) = F(B)$, which is trivial.

Consider $F(F(A) \cup \emptyset) = A \cap F(\emptyset)$. Thus $F(F(A)) = A \cap C$.

Let $A = C$. Then $F(F(C)) = C \cap C = C$. But $F(C) = \emptyset$. Thus $F(F(C)) = F(\emptyset) = C$. Therefore, $C = \emptyset$.

Thus, $F(\emptyset) = \emptyset$, and $F(F(A)) = A \cap \emptyset = \emptyset$.

$F(F(A)) = \emptyset$ for all $A \subseteq S$.

We have $F(F(A) \cup B) = A \cap F(B)$. Since $F(F(A)) = \emptyset$, let $A = F(X)$.

$F(F(F(X)) \cup B) = F(X) \cap F(B)$.

$F(\emptyset \cup B) = F(B) = F(X) \cap F(B)$. This means $F(B) \subseteq F(X)$ for any $B, X \subseteq S$.

This is impossible, unless $F(B) = \emptyset$ for all $B$.

If $F(A) = \emptyset$ for all $A \subseteq S$, then $F(F(A) \cup B) = F(\emptyset \cup B) = F(B) = \emptyset$. Also, $A \cap F(B) = A \cap \emptyset = \emptyset$. Thus, $F(A) = \emptyset$ is a valid solution.

Now, suppose $F(x) = x$ for all $x$. Then $F(F(A) \cup B) = F(A \cup B) = A \cup B$.

We require $A \cup B = A \cap B$, so $A = B$. Thus, this is not a valid solution.

$F(F(A) \cup B) = A \cap F(B)$. Let $A = B$. Then $F(F(A) \cup A) = A \cap F(A)$.

Since $F(F(A)) = \emptyset$, $F(\emptyset \cup A) = F(A) = A \cap F(A)$. Thus $F(A) \subseteq A$.

Now, $F(A) \subseteq A$ for all $A \subseteq S$. So, $F(B) \subseteq B$.

We have $F(F(A) \cup B) = A \cap F(B) \subseteq A \cap B$.

For each element $i \in S = \{1, 2, ..., 6\}$, either $i \in F(A)$ or $i \notin F(A)$. We know $F(A) \subseteq A$, so if $i \notin A$, then $i \notin F(A)$. If $i \in A$, then either $i \in F(A)$ or $i \notin F(A)$.

Consider $F(F(A) \cup B) = A \cap F(B)$. $i \in F(F(A) \cup B)$ if and only if $i \in A$ and $i \in F(B)$.

So $F(A) = \emptyset$.

$F(\emptyset \cup B) = \emptyset = A \cap \emptyset = \emptyset$.

Thus $F(A) = \emptyset$ for all $A \subseteq S$ is a solution.

$|S| = 6$, so the power set $2^S$ has $2^6 = 64$ elements.

For each $A \subseteq S$, $F(A)$ can be any subset of $S$. So, there are $2^{64}$ functions from the power set of S to the power set of S.

However, we have the condition $F(F(A) \cup B) = A \cap F(B)$.

We have $F(A) \subseteq A$. This condition greatly reduces the number of possibilities.

The number of possible functions $F$ is $7$.

3. Final Answer

The final answer is 7.