6. Let $n>1, f(x)$ be a monotonically increasing function defined on the finite set $A=$ $\{1,2, \cdots, n\}$, and for any $x$ 、 $y \in A$, we have $\frac{f(x)}{f(y)}=f(x) f(y)$. Then, ( ) .
(A) $n=2$
(B) $n=3$
(C) $n=4$
(D) $n \geqslant 5$
6. A.

For any $x \neq y, x, y \in A$, we have
$$
\frac{f(x)}{f(y)}=f(x) f(y), \frac{f(y)}{f(x)}=f(y) f(x) \text{. }
$$

Therefore, $\frac{f(x)}{f(y)}=\frac{f(y)}{f(x)}$, which means $f^{2}(x)=f^{2}(y)$.
Thus, $f^{2}(x)=c_{1}\left(c_{1}>0\right)$.
Hence, $f(x) \in\{c,-c\}(c>0)$.
If $n>2$, by the pigeonhole principle, there exist $x \neq y, x, y \in A$, such that $f(x)=f(y)$, which contradicts the fact that $f(x)$ is a strictly increasing function, so $n=2$.
Furthermore, noting that $f(x)$ is strictly increasing, we have
$$
f(1)=-c, f(2)=c \text{. }
$$

From $\frac{-c}{c}=(-c) c$, we get $c=1$.
There exists $f(x)=2 x-3, x \in\{1,2 \}$ that satisfies the condition.