how to return index of a sorted list?

You can use the python sorting functions' key parameter to sort the index array instead.

>>> s = [2, 3, 1, 4, 5, 3]
>>> sorted(range(len(s)), key=lambda k: s[k])
[2, 0, 1, 5, 3, 4]
>>> 

If you need both the sorted list and the list of indices, you could do:

L = [2,3,1,4,5]
from operator import itemgetter
indices, L_sorted = zip(*sorted(enumerate(L), key=itemgetter(1)))
list(L_sorted)
>>> [1, 2, 3, 4, 5]
list(indices)
>>> [2, 0, 1, 3, 4]

Or, for Python <2.4 (no itemgetter or sorted):

temp = [(v,i) for i,v in enumerate(L)]
temp.sort
indices, L_sorted = zip(*temp)

p.s. The zip(*iterable) idiom reverses the zip process (unzip).

Update:

To deal with your specific requirements:

"my specific need to sort a list of objects based on a property of the objects. i then need to re-order a corresponding list to match the order of the newly sorted list."

That's a long-winded way of doing it. You can achieve that with a single sort by zipping both lists together then sort using the object property as your sort key (and unzipping after).

combined = zip(obj_list, secondary_list)
zipped_sorted = sorted(combined, key=lambda x: x[0].some_obj_attribute)
obj_list, secondary_list = map(list, zip(*zipped_sorted))

Here's a simple example, using strings to represent your object. Here we use the length of the string as the key for sorting.:

str_list = ["banana", "apple", "nom", "Eeeeeeeeeeek"]
sec_list = [0.123423, 9.231, 23, 10.11001]
temp = sorted(zip(str_list, sec_list), key=lambda x: len(x[0]))
str_list, sec_list = map(list, zip(*temp))
str_list
>>> ['nom', 'apple', 'banana', 'Eeeeeeeeeeek']
sec_list
>>> [23, 9.231, 0.123423, 10.11001]

How about

l1 = [2,3,1,4,5]
l2 = [l1.index(x) for x in sorted(l1)]

you can use numpy.argsort

or you can do:

test =  [2,3,1,4,5]
idxs = list(zip(*sorted([(val, i) for i, val in enumerate(test)])))[1]

zip will rearange the list so that the first element is test and the second is the idxs.