Example 1 Try to determine all six-digit numbers $p$ with the following properties:
(1) $p, 2p, 3p, 4p, 5p, 6p$ are still six-digit numbers;
(2) the six digits of each six-digit number are still the six digits of the number $p$ (only the order of arrangement is different).
Let $p=\overline{a b c d e f}$, and denote $M=\{a, b, c, d, e, f\}$.
For any positive integer $n$, let $S(n)$ represent the sum of the digits of $n$.
From $S(3 p)=S(p)$ and $3 \mid 3 p$, we get
$3|S(3 p) \Rightarrow 3| S(p) \Rightarrow 3|p \Rightarrow 9| 3 p$
$\Rightarrow 9|S(3 p) \Rightarrow 9| S(p) \Rightarrow 9 \mid p$.
Next, we determine the set $M$.
First, note that the leading digits of the six numbers $p, 2 p, 3 p, 4 p, 5 p, 6 p$ must be distinct.

In fact, if $i p$ and $j p (1 \leqslant i < j \leqslant 6)$ have the same leading digit, then $(j-i) p$ has a leading digit of 0, meaning $(j-i) p$ has fewer than 6 digits. However, $1 \leqslant j-i \leqslant 5$, so $(j-i) p$ is one of the numbers $p, 2 p, 3 p, 4 p, 5 p$, which, by the condition, should be a six-digit number, leading to a contradiction.

Since each leading digit belongs to the set $M$, these six leading digits exactly form the set $M$. Thus, the digits of $p$ are all distinct and do not include 0.

Similarly, the last digits of these six numbers are also distinct (otherwise, if two numbers $i p$ and $j p$ have the same last digit, then the last digit of their difference is 0, and $(j-i) p$ is one of the numbers $p, 2 p, 3 p, 4 p, 5 p$, leading to a contradiction).
Therefore, the last digit of $p$ must be odd (otherwise, if the last digit of $p$ is even, then the last digit of $5 p$ is 0, which is a contradiction), and this last digit $f \neq 5$ (otherwise, the last digit of $2 p$ would be 0, which is a contradiction).

Since $6 p$ is still a six-digit number, we know that the leading digit of $p$, $a=1$ (if the leading digit of $p$, $a \geqslant 2$, then $6 p$ would be a seven-digit number, contradicting the condition). Hence, $f \neq 1$.
Thus, $f \in \{3, 7, 9\}$.
We will discuss each case separately.
If $f=3$, then the last digits of the six numbers $p, 2 p, 3 p, 4 p, 5 p, 6 p$ are $3, 6, 9, 2, 5, 8$, respectively, all belonging to the set
$$
M=\{3, 6, 9, 2, 5, 8\},
$$

These six digits should also be the six digits of $p$, but in this case, $S(p) = 33$ is not a multiple of 9, leading to a contradiction.

If $f=9$, then the last digits of the six numbers $p, 2 p, 3 p, 4 p, 5 p, 6 p$ are $9, 8, 7, 6, 5, 4$, respectively, all belonging to the set
$$
M=\{9, 8, 7, 6, 5, 4\},
$$

These six digits should also be the six digits of $p$, but in this case, $S(p) = 39$, which is not a multiple of 9, leading to a contradiction.
Therefore, only $f=7$.
Thus, the last digits of the six numbers $p, 2 p, 3 p, 4 p, 5 p, 6 p$ are $7, 4, 1, 8, 5, 2$.
Hence, the set $M=\{7, 4, 1, 8, 5, 2\}$.
We already have $a=1$ and $f=7$, so $p=\overline{1 b c d e 7}$.
Thus, $b \in \{2, 4, 5, 8\}$.
The leading digits of $p, 2 p, 3 p, 4 p, 5 p, 6 p$ should be the increasing sequence of numbers in $M$, i.e., $1, 2, 4, 5, 7, 8$.

If $b \geqslant 5$, then the leading digit of $2 p$ will be greater than 2, leading to a contradiction. Hence, $b \in \{2, 4\}$.

If $b=2$, then $p \leqslant 128547$, and in this case, the leading digit of $5 p$ is no more than 6, leading to a contradiction.
Thus, only $b=4$, giving $p=\overline{14 c d e 7}$.
Thus, $e \in \{2, 5, 8\}$.
If $e=2$, i.e., $p=\overline{14 c d 27}$, then the second last digit (tens place) of $4 p$ is 0, and $0 \notin M$;

If $e=8$, i.e., $p=\overline{14 c d 87}$, then the second last digit (tens place) of $3 p$ is 6, and $6 \notin M$.

Therefore, only $e=5$, giving $p=\overline{14 c d 57}$, where $\{c, d\}=\{2, 8\}$. In this case, only $p=148257$ or $p=142857$ are possible.

If $p=148257$, then $2 p=296514$, and the digits $6, 9 \notin M$.
Thus, the only possible six-digit number $p$ in this problem is $p=142857$.
It is easy to verify that it satisfies the conditions of the problem.