Let $$A = \{1, 2, 3, 5, 8, 9\}$$. Then the number of possible functions $$f : A \to A$$ such that $$f(m \cdot n) = f(m) \cdot f(n)$$ for every $$m, n \in A$$ with $$m \cdot n \in A$$ is equal to

We need to find the number of functions $$f : A \to A$$ such that $$f(m \cdot n) = f(m) \cdot f(n)$$ for every $$m, n \in A$$ with $$m \cdot n \in A$$, where $$A = \{1, 2, 3, 5, 8, 9\}$$.

We begin by identifying all pairs $$(m, n)$$ with $$m, n \in A$$ and $$m \cdot n \in A$$. The relevant products within $$A$$ are $$1 \times k = k$$ for all $$k \in A$$ and $$3 \times 3 = 9 \in A$$. All other products (for example, $$2 \times 3 = 6$$, $$2 \times 5 = 10$$, $$3 \times 5 = 15$$, etc.) fall outside $$A$$.

From $$f(1 \cdot k) = f(1) \cdot f(k)$$ for all $$k \in A$$ we obtain $$f(k) = f(1) \cdot f(k)$$. Since $$f(k) \ge 1$$ for all $$k$$ (because $$f : A \to A$$ and $$0 \notin A$$), we must have $$f(1) = 1$$.

Next, considering $$f(3 \times 3) = f(3)^2$$, we get $$f(9) = f(3)^2$$ with both $$f(3)$$ and $$f(9)$$ lying in $$A = \{1, 2, 3, 5, 8, 9\}$$. Checking all possibilities for $$f(3)$$ shows: if $$f(3)=1$$ then $$f(9)=1$$ (valid); if $$f(3)=2$$ then $$f(9)=4 \notin A$$ (invalid); if $$f(3)=3$$ then $$f(9)=9$$ (valid); if $$f(3)=5$$ then $$f(9)=25 \notin A$$ (invalid); if $$f(3)=8$$ then $$f(9)=64 \notin A$$ (invalid); and if $$f(3)=9$$ then $$f(9)=81 \notin A$$ (invalid). Therefore $$f(3) \in \{1, 3\}$$, giving 2 choices, each determining $$f(9)$$.

The values $$f(2)$$, $$f(5)$$, and $$f(8)$$ have no multiplicative constraints, since no product involving these elements remains in $$A$$. Hence each of these can be chosen freely as any of the 6 elements of $$A$$.

Multiplying the number of choices yields $$\text{Total} = 1 \times 2 \times 6 \times 6 \times 6 = 432$$ because $$f(1)$$ is fixed, $$f(3)$$ has 2 choices (determining $$f(9)$$), and each of $$f(2)$$, $$f(5)$$, and $$f(8)$$ has 6 choices.

Thus the correct answer is \boxed{432}.