Example 1 Let $m=2^{l}(l \geqslant 3), a=5$. Find $d_{0}=\delta_{m}(a)$.
Given $\varphi\left(2^{l}\right)=2^{l-1}$ and $d_{0} \mid \varphi\left(2^{l}\right)$, we know that $d_{0}=2^{k}, 0 \leqslant k \leqslant l-1$. First, we prove that for any $a, 2 \nmid a$, it must be true that
$$a^{2^{2-2}} \equiv 1\left(\bmod 2^{l}\right)$$

We use induction on $l$ to prove equation (20). Let $a=2 t+1$. When $l=3$,
$$a^{2}=4 t(t+1)+1 \equiv 1\left(\bmod 2^{3}\right),$$

so equation (20) holds. Assume equation (20) holds for $l=n(\geqslant 3)$. When $l=n+1$, by
$$a^{2^{n-1}}-1=\left(a^{2^{n-2}}-1\right)\left(a^{2^{n-2}}+1\right)$$

and the induction hypothesis, we have
$$a^{2^{n-1}}-1 \equiv 0\left(\bmod 2^{n+1}\right),$$

which means equation (20) holds for $l=n+1$. This proves that equation (20) holds for any $l \geqslant 3$. Therefore, for any $a(2 \nmid a)$, the corresponding $d_{0}=2^{k}$ must satisfy $0 \leqslant k \leqslant l-2$. Next, we find $d_{0}$ for $a=5$. By
$$5^{2^{0}} \equiv 5^{1} \not \equiv 1\left(\bmod 2^{3}\right)$$

and equation (20) with $a=5, l=3$, we conclude that when $l=3$, $d_{0}=2^{3-2}=2$. By
$$5^{2^{1}} \equiv 25 \not \equiv 1\left(\bmod 2^{4}\right)$$

and equation (20) with $a=5, l=4$, we conclude that when $l=4$, $d_{0}=2^{4-2}=2^{2}$. We now prove that for any $l \geqslant 3$, it must be true that
$$5^{2^{l-3}} \not \equiv 1\left(\bmod 2^{l}\right)$$

We use induction on $l$. When $l=3$, equation (22) holds by equation (21). Assume equation (22) holds for $l=n$ $(\geqslant 3)$. Since when $l \geqslant 3$,
$$5^{2^{l-3}} \equiv 1\left(\bmod 2^{l-1}\right)$$

which can be verified directly when $l=3$ and is equation (20) when $l>3$. By the induction hypothesis and equation (23), we have
$$5^{2^{n-3}}=1+s \cdot 2^{n-1}, \quad 2 \nmid s,$$

and thus (note $n \geqslant 3$)
$$5^{2^{n-2}}=1+s\left(1+s \cdot 2^{n-2}\right) 2^{n}, \quad 2 \nmid s\left(1+s \cdot 2^{n-2}\right).$$

This proves that equation (22) holds for $l=n+1$. Therefore, equation (22) holds for any $l \geqslant 3$. This implies (why)
$$5^{2^{j}} \not \equiv 1\left(\bmod 2^{l}\right), \quad 0 \leqslant j \leqslant l-3$$

By this and equation (20), we conclude that when $a=5$, the corresponding $d_{0}=2^{l-2}$. From Theorem 4, when $l \geqslant 3$,
$$5^{0}=1,5^{1}, 5^{2}, 5^{3}, \cdots, 5^{2^{l-2}-1}$$

these $2^{l-2}$ numbers are pairwise incongruent modulo $2^{l}$.