5. Given that $f(n)$ is a function defined on the set of natural numbers $\mathbf{N}$, and satisfies
$$
\begin{array}{l}
f(1)=f(2)=1, f(3 n)=3 f(n)-2, f(3 n+1)=3 f(n)+1, f(3 n+2)= \\
3 f(n)+4(n \in \mathbf{N}) .
\end{array}
$$

Determine the largest positive integer $n$ less than or equal to 1992 for which $f(n)=n$ holds.
5. Direct calculation and observation of patterns
\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
\hline$f(n)$ & 1 & 1 & 1 & 4 & 7 & 1 & 4 & 7 & 1 & 4 & 7 & 10 & 13 & 16 \\
\hline$n$ & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 \\
\hline$f(n)$ & 19 & 22 & 25 & 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 & 25 & 1 & 4 \\
\hline
\end{tabular}

Therefore, we can conjecture
Next, we use mathematical induction to prove for $m$:
When $m=1$, $a=1$ or $2$, $f(a \cdot 3+0)=3 f(a)-2=1=3 \times 0+1$,
$f(a \cdot 3+1)=3 f(a)+1=4=3 \times 1+1$,
$f(a \cdot 3+2)=3 f(a)+4=7=3 \times 2+1$,

That is, when $m=1$, formula $(*)$ holds.
Assume that when $m=r$, formula $(*)$ holds. We need to prove the case for $m=r+1$.
(i) If $k=3 t$, then
$f\left(a \cdot 3^{r+1}+k\right)=f\left(a \cdot 3^{r+1}+3 t\right)=3 f\left(a \cdot 3^{r}+t\right)-2=3(3 t+1)-2=3(3 t)+1=3 k+1$.
(ii) If $k=3 t+1$, then
$f\left(a \cdot 3^{r+1}+k\right)=f\left(a \cdot 3^{r+1}+3 t+1\right)=3 f\left(a \cdot 3^{r}+t\right)+1=3(3 t+1)+1=3 k+1$.
(iii) If $k=3 t+2$, then
$f\left(a \cdot 3^{r+1} k\right)=f\left(a \cdot 3^{r+1}+3 t+2\right)=3 f\left(a \cdot 3^{r}+t\right)+4=3(3 t+1)+4=3(3 t+2)+1=3 k+1$,
That is, when $m=r+1$, formula $(*)$ also holds. This proves that formula $(*)$ is true.
By $f(n)=n$, i.e., $f\left(a \cdot 3^{m}+k\right)=a \cdot 3^{m}+k$,
Also, $3 k+1=a \cdot 3^{m}+k \Rightarrow k=\frac{1}{2}\left(a \cdot 3^{m}-1\right)$.
And $k=\frac{1}{2}\left(a \cdot 3^{m}-1\right)<3^{m} \Rightarrow a=1$.
Therefore, when $n \geqslant 3$, $f(n)=n \Leftrightarrow n=3^{m}+\frac{1}{2}\left(3^{m}-1\right)=\frac{1}{2}\left(3^{m+1}-1\right)$. Also, $f(1)=1, f(2) \neq 2$, so when $m=0$, $n=\frac{1}{2}(3-1)=1$. Thus,
$f(n)=n \Leftrightarrow n=\frac{1}{2}\left(3^{m+1}-1\right), m=0,1,2, \cdots$.
By $\frac{1}{2}\left(3^{m+1}-1\right) \leqslant 1992 \Rightarrow 3^{m} \leqslant \frac{1}{3} \times 3985<3^{7}$,
Hence $m<7$, so $m_{\max }=6, n_{\max }=\frac{1}{2}\left(3^{6+1}-1\right)=1093$.
Another solution: Suppose $n$ is represented in base 3 as
$n=\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)_{3}, a_{i}=0,1,2(0 \leqslant i \leqslant k-1), a_{k}=1$ or 2,
Then $f\left(\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)_{3}\right)=\left(a_{k-1} a_{k-2} \cdots a_{1} a_{0} 1\right)_{3}$.
We use mathematical induction to prove formula $(*)$.
$$
\begin{array}{l}
f(1)=f\left((1)_{3}\right)=1=(1)_{3}, \\
f(2)=f\left((2)_{3}\right)=1=(1)_{3}, \\
f(3)=f\left((10)_{3}\right)=3 f(1)-2=1=(01)_{3},
\end{array}
$$

That is, when $n=1,2,3$, formula $(*)$ holds. Assume that when $n<l$, conclusion $(*)$ holds. We need to prove the case for $n=l$. Let $m=\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)$, then (i) $l=3 m(m \in \mathbf{N})$, then $l\left(a_{k} a_{k-1} \cdots a_{1} a_{0} 0\right)_{3}$,
$f(l)=f(3 m)=3 f(m)-2=3 \times\left(a_{k-1} \cdots a_{1} a_{0} 1\right)_{3}-2=\left(a_{k-1} \cdots a_{0} 01\right)_{3}$.
(ii) $l=3 m+1(m \in \mathbf{N})$, then $l=\left(a_{k} a_{k-1} \cdots a_{1} a_{0} 1\right)_{3}$,
$f(l)=3 f(m)+1=3 \times\left(a_{k-1} \cdots a_{1} a_{0}\right)+1=\left(a_{k-1} \cdots a_{1} a_{0} 11\right)_{3}$.
(iii) $l=3 m+2(m \dot{\in} \mathbf{N})$, then $l=\left(a_{k} a_{k-1} \cdots a_{1} a_{0} 2\right)_{3}$,
$f(l)=3 f(m)+4=3 \times\left(a_{k-1} \cdots a_{1} a_{0} 1\right)_{3}+(11)_{3}=\left(a_{k-1} \cdots a_{1} a_{0} 21\right)_{3}$.
Combining (i), (ii), and (iii), we know that for all $n \in \mathbf{N}$, formula $(*)$ holds.
By $f(n)=n$, let $n=\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)_{3}$, we get $\left(a_{k-1} a_{k-2} \cdots a_{1} a_{0} 1\right)_{3}=\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)_{3}$, comparing the digits in each position, we get $a_{k}=a_{k-1}=\cdots=a_{1}=a_{0}=1$,
That is, $f(n)=n \Leftrightarrow n=(11 \cdots 1)_{3}$.
Also, $(100000)_{3}=3^{6}<1992<3^{7}=(1000000)_{3}$,
Therefore, the largest positive integer $n$ such that $f(n)=n$ holds for all positive integers less than or equal to 1992 is
$$
n_{\max }=(111111)_{3}=3^{6}+3^{5}+3^{4}+3^{3}+3^{2}+3^{1}+3^{0}=\left(3^{7}-1\right) /(3-1)=1093 .
$$