Example 2. Find all natural numbers with the following property: its unit digit is 6, and when its unit digit is moved to the front of the number, the resulting new number is four times the original number.
Let the required number be $a_{n} a_{n-1} \cdots \cdots a_{2} a_{1} 6$. According to the problem, we have $\overline{6 a_{n} a_{n-1} \cdots a_{2} a_{1}}=4 \overline{a_{n} a_{n-1} \cdots a_{2} a_{1} 6}$, where $0 \leqslant a_{i} \leqslant 9, a_{i} \in N$.
That is 6
$$
\begin{array}{l}
6 \cdot 10^{n}+a_{n} \cdot 10^{n-1}+\cdots+a_{2} \cdot 10+a_{1} \\
=4\left(a_{n} \cdot 10^{n}+a_{n-1} \cdot 10^{n-1}+\cdots+a_{2} \cdot 10^{2}\right. \\
\left.\quad+a_{1} \cdot 10+6\right) .
\end{array}
$$

Thus, $\left(6-4 a_{n}\right) \cdot 10^{n}+\left(a_{n}-4 a_{n}-1\right) \cdot 10^{n-1}+\cdots+$ $\left(a_{2}-4 a_{1}-2\right) \cdot 10+a_{1}-4=0$, and from $6-4 a_{n} > 0$ we know $a_{n}<\frac{3}{2}$. Therefore, $a_{n}=1$.
Hence, the required number must be 153846, 153846153846, $153846153846153846, \cdots$.