Example 6 Suppose it is known that 2 is a primitive root of $p=13$, try to find all the primitive roots of $p$.
Since $\varphi(p-1)=\varphi(12)=\varphi(4) \varphi(3)=2 \cdot 2=4$, there are exactly 4 primitive roots of 13. Among the 12 natural numbers $1,2, \cdots, p-1=12$, the numbers that are coprime with 12 are exactly the following 4: $1,5,7,11$. Therefore, the following 4 numbers $2^{1} \equiv 2,2^{5} \equiv 6,2^{7} \equiv 11,2^{11} \equiv 7(\bmod 13)$ are precisely all the primitive roots of 13.

This method is also fully applicable to finding all $l$-th power residues of $p$ $(l \mid(p-1))$. Since $p-1=12$ has the divisors $1,2,3,4,6,12$, we know that modulo 13 there are $\varphi(1)=1$ first-order elements, $\varphi(2)=1$ second-order elements, $\varphi(3)=2$ third-order elements, $\varphi(4)=2$ fourth-order elements, $\varphi(6)=2$ sixth-order elements, and $\varphi(12)=4$ primitive roots. Clearly, 1 is a first-order element; $-1 \equiv 12$ is a second-order element; since $3^{3} \equiv 1$ and $3^{2} \equiv 1(\bmod 13)$, 3 is a third-order element. Noting that among $1,2,3$ the numbers coprime with 3 are 1 and 2, the complete set of third-order elements modulo 13 is $3^{1}$ and $3^{2} \equiv 9(\bmod 13)$. Since $5^{2} \equiv-1,5^{3} \equiv-5$, and $5^{4} \equiv 1(\bmod 13)$, 5 is a fourth-order element of 13. Noting that among $1,2,3,4$ the numbers coprime with 4 are 1 and 3, the complete set of fourth-order elements modulo 13 is the following two: $5^{1}, 5^{3}=8(\bmod 13)$. From the above calculations, it is clear that the remaining 4 and 10 must be precisely all the sixth-order elements modulo 13.