6.1.3 * Find all polynomials $f(x)$ such that
$$
x f(x-1)=(x-15) f(x) .
$$
From the condition, we know that $x \mid f(x)$. Let $f(x) = x f_{1}(x)$, and substitute into equation (1). After canceling $x$ on both sides, we get
$$
(x-1) f_{1}(x-1) = (x-15) f_{1}(x).
$$

Similarly, we know that $x-1 \mid f_{1}(x)$. Continuing this process, we can set $f(x) = x(x-1) \cdots (x-14) f_{15}(x)$, and substitute into equation (1) to get
$$
x(x-1) \cdots (x-15) f_{15}(x-1) = (x-15) x \cdots (x-14) f_{15}(x).
$$

Therefore, $f_{15}(x-1) - f_{15}(x)$. If $f_{15}(x)$ is not a constant, and its roots are $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$, then $\alpha_{1} + 1, \alpha_{2} + 1, \cdots, \alpha_{k} + 1$ are also its roots. However, $\alpha_{1} + \alpha_{2} + \cdots + \alpha_{k} \neq (\alpha_{1} + 1) + (\alpha_{2} + 1) + \cdots + (\alpha_{k} + 1)$, which is a contradiction! Therefore, $f_{15}(x)$ is a constant.
Thus, $f(x) = c x(x-1) \cdots (x-14)$, where $c$ is any constant.