Example 7 In the real-coefficient polynomial $p(x)=x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, for any root $\alpha$, $\frac{1}{\alpha}, 1-\alpha$ are also roots of $p(x)$. Find $p(x)$.
Analyzing $p(x)$ as a 5th-degree polynomial, it can have at most 5 roots. According to the problem, if $p(\alpha)=0$, then $p\left(\frac{1}{\alpha}\right)=0$, $p(1-\alpha)=0$, and when $\alpha \neq 1$, $p\left(\frac{1}{1-\alpha}\right)=0$, $p\left(1-\frac{1}{\alpha}\right)=0$, and $p\left(1-\frac{1}{1-\alpha}\right)=0$. Therefore, when $\alpha \neq 1$, $\alpha, \frac{1}{\alpha}, 1-\alpha, \frac{1}{1-\alpha}, 1-\frac{1}{\alpha}, 1-\frac{1}{1-\alpha}$ are all roots of $p(x)$, from which the value of $\alpha$ can be determined.

Solution: According to the problem, if $\alpha$ is a root of $p(x)=0$, and when $\alpha \neq 1$, $\frac{1}{\alpha}, 1-\alpha, \frac{1}{1-\alpha}, 1-\frac{1}{\alpha}, 1-\frac{1}{1-\alpha}$ are also roots of $p(x)$. Since $p(x)$ is a 5th-degree polynomial, it can have at most 5 roots, so among these 6 roots, there must be identical values.

Let $\alpha=\frac{1}{\alpha}$, or $\alpha=1-\alpha$, or $\alpha=\frac{1}{1-\alpha}$, or $\alpha=1-\frac{1}{\alpha}$, or $\alpha=1-\frac{1}{1-\alpha}$, then $\alpha$ can be one of $-1, \frac{1}{2}, 2, \frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}$. Therefore, we have
$$
\begin{array}{l}
p(x)=(x+1)\left(x-\frac{1}{2}\right)(x-2)\left(x^{2}-x+1\right), \\
p(x)=(x+1)^{3}\left(x-\frac{1}{2}\right)(x-2), \\
p(x)=(x+1)\left(x-\frac{1}{2}\right)^{3}(x-2), \\
p(x)=(x+1)\left(x-\frac{1}{2}\right)(x-2)^{3}, \\
p(x)=(x+1)^{2}\left(x-\frac{1}{2}\right)^{2}(x-2), \\
p(x)=(x+1)\left(x-\frac{1}{2}\right)^{2}(x-2)^{2}, \\
p(x)=(x+1)^{2}\left(x-\frac{1}{2}\right)(x-2)^{2},
\end{array}
$$

all satisfy the conditions.
In particular, when $\alpha=1$, $p(x)=(x-1)^{5}$ also satisfies the conditions.