Example 8 Find the odd prime $p$ that satisfies the following condition: there exists a permutation $b_{1}, b_{2}, \cdots, b_{p-1}$ of $1,2, \cdots, p-1$, such that $1^{b_{1}}, 2^{b_{2}}, \cdots,(p-1)^{b_{p-1}}$ forms a reduced residue system modulo $p$.
Let $p$ be an odd prime satisfying the condition, and take $g$ as a primitive root of $p$. Since $g, g^{2}, \cdots, g^{p-1}$ form a reduced residue system modulo $p$, for each $i=1,2, \cdots, p-1$, $i$ can be represented as $g^{a_{i}}$ modulo $p$, where $a_{1}, a_{2}, \cdots, a_{p-1} \in \mathbf{Z}_{+}$, and by property 5 (2), $a_{1}, a_{2}, \cdots, a_{p-1}$ form a complete residue system modulo $p-1$.

Therefore, the problem is transformed into: If $a_{1}, a_{2}, \cdots, a_{p-1}$ form a complete residue system modulo $p-1$, can we find another complete residue system $b_{1}, b_{2}, \cdots, b_{p-1}$ modulo $p-1$, such that $g^{a_{1} b_{1}}, g^{a_{2} b_{2}}, \cdots, g^{a_{p-1} b_{p-1}}$ form a reduced residue system modulo $p$? In other words, do $a_{i} b_{i}(1 \leqslant i \leqslant p-1)$ form a complete residue system modulo $p-1$?

By Example 7, we know that the conclusion is affirmative if and only if $p-1 \leqslant 2$, i.e., $p=3$.

In this problem, the corresponding example is $b_{1}=2, b_{2}=1$. Therefore, $p=3$ is the only odd prime that satisfies the condition.