$6 \cdot 137$ Given a function $f(n)$ on the set of positive integers, satisfying the following conditions:
$f(n)=n-12$, when $n>2000$;
$f(n)=f(f(n+16))$, when $n \leqslant 2000$.
(1) Find $f(n)$.
(2) Find all solutions to the equation $f(n)=n$.
［Solution］(1) From the given information,
$$
\begin{array}{l}
f(2000)=f(f(2016))=f(2004)=1992, \\
f(1999)=f(f(2015))=f(2003)=1991, \\
f(1998)=f(f(2014))=f(2002)=1990 \\
f(1997)=f(f(2013))=f(2001)=1989 \\
f(1996)=f(f(2012))=f(2000)=1992 \\
f(1995)=f(f(2011))=f(1999)=1991, \\
f(1994)=f(f(2010))=f(1998)=1990 \\
f(1993)=f(f(2009))=f(1997)=1989 \\
f(1992)=f(f(2008))=f(1996)=1992 \\
f(1991)=f(f(2007))=f(1995)=1991, \\
f(1990)=f(f(2006))=f(1994)=1990 \\
f(1989)=f(f(2005))=f(1993)=1989 \\
f(1988)=f(f(2004))=f(1992)=1992 \\
f(1987)=f(f(2003))=f(1991)=1991, \\
f(1986)=f(f(2002))=f(1990)=1990 \\
f(1985)=f(f(2001))=f(1989)=1989 \\
f(1984)=f(f(2000))=f(1992)=1992 \\
f(1983)=f(f(1999))=f(1991)=1991, \\
f(1982)=f(f(1998))=f(1990)=1990 \\
f(1981)=f(f(1997))=f(1989)=1989
\end{array}
$$

Thus, we conjecture: For non-negative integers $k \leqslant 499, m \in\{0,1,2,3\}$, we have
$$
f(2000-4 k-m)=1992-m .
$$

We will prove this conjecture using mathematical induction.
When $k=0,1,2,3$, for any $m \in\{0,1,2,3\}$, we have already verified that the conjecture holds.
Assume that the conjecture holds for $k \leqslant t(t \geqslant 3)$.
Then for $k=t+1$,
$$
\begin{aligned}
1 & \leqslant[2000-4(t+1)-m]+16 \\
& =2000-4(t-3)-m \\
& \leqslant 2000,
\end{aligned}
$$

Therefore, by the given conditions and the induction hypothesis, we have
$$
\begin{aligned}
f(2000-4(t+1)-m) & =f(f(2000-4(t-3)-m)) \\
& =f(1992-m) \\
& =f(2000-4 \cdot 2-m) \\
& =1992-m .
\end{aligned}
$$

Thus, the conjecture holds for any non-negative integer $k \leqslant 499$ and $m \in\{0,1,2,3\}$. Therefore, we have
$$
f(n)=\left\{\begin{array}{l}
n-12, \text { when } n>2000 \text {; } \\
1992-m, \text { when } n=2000-4 K-m \text {; }
\end{array}\right.
$$

Here $K$ is a non-negative integer, $m \in\{0,1,2,3\}$, and $K \leqslant 499$.
(2) If $f(n)=n$, then $n \leqslant 2000$ must hold.

Let
$$
n=2000-4 k-m,
$$

then
$$
1992-m=2000-4 k-m,
$$

thus
$$
k=2,
$$

therefore
$$
n=1992-m, m \in\{0,1,2,3\},
$$

so the integers $n$ that satisfy $f(n)=n$ are $1992,1991,1990,1989$.