Problem 943

To prove: The problem is to prove that $f(n)$ is a fixed point for any natural number $n (n \geqslant 2)$ under the given conditions. In fact,
$$\begin{array}{l} f(3) f(7)=f(21) \\ f(2)=2, \text{ so, } f(3)=3. \end{array}$$
We will prove this by contradiction.
If the original proposition is not true, assume the smallest positive integer $n$ for which $f(n) \neq n$ is $n_{0} \geqslant 4$, then
$$f\left(n_{0}\right)>f\left(n_{0}-1\right)=n_{0}-1.$$

Therefore, it can only be that $f\left(n_{0}\right)>n_{0}$.
Also, $\{f(n)\}$ is strictly increasing, so when $n \geqslant n_{0}$, we have
$$f(n)>n.$$

We will discuss this in two cases:
(1) When $n_{0}$ is odd, 2 and $n_{0}-2$ are coprime, then
$$\begin{array}{l} f\left[2\left(n_{0}-2\right)\right]=f(2) f\left(n_{0}-2\right) \\ =2\left(n_{0}-2\right). \end{array}$$

Since $n_{0} \geqslant 4$, then $2\left(n_{0}-2\right) \geqslant n_{0}$. Thus, equation (2) contradicts equation (1).
(2) When $n_{0}$ is even, 2 and $n_{0}-1$ are coprime, similarly, we have
$$\begin{array}{l} f\left[2\left(n_{0}-1\right)\right]=f(2) f\left(n_{0}-1\right) \\ =2\left(n_{0}-1\right). \end{array}$$

Since $n_{0} \geqslant 4$, then $2\left(n_{0}-1\right) \geqslant n_{0}$. Thus, equation (3) contradicts equation (1).
In summary, equation (1) does not hold. Therefore, $f(n)=n$.