(6) The function $f(x)$ is a monotonically increasing function on $(0,+\infty)$, when $n \in \mathbf{N}^{*}$, $f(n) \in$ $\mathbf{N}^{*}$, and $f(f(n))=3 n$, then the value of $f(1)$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6 (by the method of elimination) Let $n=1$, then we get $f(f(1))=3$. If $f(1)=1$, then
$$
f(f(1))=f(1)=3,
$$

which contradicts $f(1)=1$;
If $f(1)=3$, then
$$
f(f(1))=f(3)=3,
$$

which contradicts the fact that $f(x)$ is monotonically increasing on $(0,+\infty)$;
If $f(1)=4$, then
$$
f(f(1))=f(4)=3,
$$

which also contradicts the fact that $f(x)$ is monotonically increasing on $(0,+\infty)$. Therefore, the answer is B.