Project Euler Problem 52: Permuted multiples
Python lacks C-style techniques like side-effect assignments (while (++x) { … }) and counting for-loops (for (int x = 125874; ; x++) { … }).  When you find yourself needing a counting loop in Python, a better formulation is nearly always available using either xrange(), itertools, or a generator.  In this case, I recommend itertools.count().

There's nothing in the problem statement that implies that the answer exceeds 125874.  The only logical limit I can see is that the answer needs to have at least six digits.  (The problem is probably designed to make you think about the decimal representation of \$\frac{1}{7}\$.)

Furthermore, manually writing all of the tests for 2, 3, 4, 5, and 6 is poor style.

There is no need to transform each set into a sorted list.  Skipping that transformation nearly doubles the speed of the code.

from itertools import count

for n in count(100000):
    digits = set(str(n))
    if all(set(str(mult * n)) == digits for mult in xrange(2, 7)):
        print n
        break