Observation is insight

Although the Brute force approach can effectively solve the problem, exploring alternative methodologies can offer valuable insights. In this case, several pertinent observations can be made:

1. The number must be a multiple of \( 9 \), because every number are congruent to the sum of their digits modulo \( 9 \), thus \( x \equiv s \pmod{9} \), where \( s \) is the sum of digits of \( x \). Since \( 2x \equiv s \pmod{9} \), it follows that \( 2x - x \equiv s - s \pmod{9} \), which implies that \( x \equiv 0 \pmod{9} \).
2. The first digit of the number must be \( 1 \), otherwise \( 2x \) would have one more digit than \( x \).
3. According to the previous observation, \( 2x \) must start with \( 2 \) or \( 3 \), \( 3x \) must start with \( 3 \) or \( 4 \) and so on. Therefore, the number must at least contain 6 digits to satisfy to contain digits from \( 1 \) to \( 6 \).
4. The number must contain a \( 0 \) or a \( 5 \) because \( 5x \) is obviously a multiple of \( 5 \).

It's actually possible to continue with the observation and find the solution by hand.

From solution2.py:

def permuted_multiples():
    for i in itertools.count(100008, 9):  # Observation 1 and 4
        s = str(i)
        if int(s[0]) != 1 or all(d not in s for d in "05"):  # Observation 2 and 3
            continue
        if all(sorted(s) == sorted(str(i * j)) for j in range(2, 7)):
            return i

It is worth noting that the solution is trivial if one know the property of \frac{1}{7} in decimal representation.