5. Given that $f(x)$ is a linear function, $f[f(1)]=-1$, and the image of $f(x)$ symmetric to the line $x-y=0$ is $C$. If the point $\left(n, \frac{a_{n+1}}{a_{n}}\right)\left(n \in \mathbf{N}^{\cdot}\right)$ lies on the curve $C$, and $a_{1}=1, \frac{a_{n+1}}{a_{n}}-\frac{a_{n}}{a_{n-1}}=1(n \geqslant 2)$.
(1) Find the analytical expression of $f(x)$ and the equation of the curve $C$;
(2) Let $S_{n}=\frac{a_{1}}{3!}+\frac{a_{2}}{4!}+\frac{a_{3}}{5!}+\cdots+\frac{a_{n}}{(n+2)!}$, for all $n \in \mathbf{N}^{*}$, if $S_{n}>m$ always holds, find the maximum value of the natural number $m$.
5. Solution (1) Let $f(x)=k x+b(k \neq 0), \therefore f[f(1)]=k^{2}+k b+b=-1$. (1) Since the image of $f(x)$ is symmetric about the line $x-y=0$, the curve $C$ is: $f^{-1}(x)=\frac{x}{k}-\frac{b}{k}, \therefore f^{-1}(n)=\frac{n}{k}-\frac{b}{k}, f^{-1}(n-1)=\frac{n-1}{k}-\frac{b}{k}, f^{-1}(n)-f^{-1}(n-1)=\frac{1}{k}$, and the point $\left(n, \frac{a_{n-1}}{a_{n}}\right)\left(n \in \mathbf{N}^{\cdot}\right)$ lies on the curve $C$,
$f^{-1}(n)=\frac{a_{n-1}}{a_{n}}$ (2), $f^{-1}(n-1)=\frac{a_{n}}{a_{n-1}}, \therefore f^{-1}(n)-f^{-1}(n-1)=\frac{a_{n+1}}{a_{n}}-\frac{a_{n}}{a_{n-1}}=1$, we get $k=1, b=-1$, thus, $f(x)=x-1$, and the curve C: $y=x+1$.
(2) From (2) $f^{-1}(n)=\frac{a_{n-1}}{a_{n}}, \therefore \frac{a_{n-1}}{a_{n}}=n+1, \therefore \frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \cdots \cdot \frac{a_{3}}{a_{2}} \cdot \frac{a_{2}}{a_{1}}=n(n-1) \cdots 3 \cdot 2=n$ !

From $a_{1}=1$, we get $a_{n}=n!$, and $\frac{a_{n}}{(n+2)!}=\frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)}=\frac{1}{n+1}-\frac{1}{n+2}$.
Thus, $S_{n}=\frac{a_{1}!}{3!}+\frac{a_{2}!}{4!}+\frac{a_{3}!}{5!}+\cdots+\frac{a_{n}!}{(n+2)!}=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$ $\frac{1}{2}-\frac{1}{n+2}$, and $0<\frac{1}{n+2} \leqslant \frac{1}{3}, \frac{1}{6} \leqslant \frac{1}{2}-\frac{1}{n+2}<\frac{1}{2}$, so the minimum value of $S_{n}$ is $\frac{1}{6}$.

Therefore, $m<\frac{1}{6}$, the maximum value of the natural number $m$ is 0.