Example 3 Solve the congruence equation $x^{11} \equiv 27\left(\bmod 2^{6}\right)$.
Solve Table 4 of Example 2 in §3 to get $\gamma^{(-1)}(27)=1, \gamma^{(0)}(27)=9$. Therefore, we need to solve two linear congruences:
$$\begin{array}{ll}
11 u \equiv 1(\bmod 2), & 0 \leqslant u<1, \\
11 v \equiv 9\left(\bmod 2^{4}\right), & 0 \leqslant v<2^{4} .
\end{array}$$

It is easy to solve and get $u=1, v=11$. Thus, the $x$ given by equation (11) is the solution to the original equation. By checking Table 3 of Example 2 in §3, we get the solution $x \equiv-29\left(\bmod 2^{6}\right)$.