Example 12 The function $f: \mathrm{N}^{+} \rightarrow \mathrm{N}^{+}$ satisfies
(1) $f(n+1)>f(n)$,
(2) $f[f(n)]=3 n$.

Find $f(1992)$.
Let's first find the expression for $f$, and then calculate $f(1992)$.
First, we have $f(1)=2, f(2)=3$.
If $f(1)=1$, then $f[f(1)]=f(1)=1 \neq 3$, which contradicts (2).
If $f(1) \geqslant 3$, by condition (1), $f(3) \geqslant f(2)+1 \geqslant f(1)+2 \geqslant 5$. Hence,
$$
f[f(1)] \geqslant f(3) \geqslant 5 \neq 3.
$$

This contradicts $f[f(1)]=3$. Therefore,
$$
\begin{aligned}
f(1) & =2, f(2)=f[f(1)]=3. \\
f(3 n) & =f(f(f(n)))=f^{(2)}(f(n)) \\
& =3 f(n),
\end{aligned}
$$

Because
$$
\begin{aligned}
f(3 n) & =f(f(f(n)))=f^{(2)}(f(n)) \\
& =3 f(n),
\end{aligned}
$$

That is,
$$
f(3 n)=3 f(n).
$$

So, from (1) and (2), we immediately get (which can be easily proved by induction)
$$
\begin{array}{l}
f\left(3^{n}\right)=2 \cdot 3^{n}, \\
f\left(2 \cdot 3^{n}\right)=3^{n+1}, \quad n=0,1,2, \cdots.
\end{array}
$$

Notice that there are exactly $3^{n}-1$ natural numbers between $2 \cdot 3^{n}$ and $3^{n+1}$, and also exactly $3^{n}-1$ natural numbers between $3^{n}$ and $2 \cdot 3^{n}$. By condition (1) (i.e., $f$ is a strictly increasing function), we have
$$
\begin{array}{l}
f\left(3^{n}+m\right)=2 \cdot 3^{n}+m \\
0 \leqslant m \leqslant 3^{n}, n=0,1,2, \cdots
\end{array}
$$

From (3), we get
$$
f\left(2 \cdot 3^{n}+m\right)=f\left(f\left(3^{n}+m\right)\right)=3\left(3^{n}+m\right).
$$

Thus, we have
$$
f(n)=\left\{\begin{array}{l}
2 \cdot 3^{k}+m, \text { if } n=3^{k}+m, 0 \leqslant m \leqslant 3^{k}, \\
3\left(3^{k}+m\right), \text { if } n=2 \cdot 3^{k}+m, 0 \leqslant m \leqslant 3^{k}.
\end{array}\right.
$$

Since $1992=2 \cdot 3^{6}+534$, we have
$$
f(1992)=3\left(3^{6}+534\right)=3789.
$$