Sort or remove elements from corresponding list in same way as in reference list

Try this:

[i for j, i in sorted(zip(listA, listB))]

Output:

listA = [7, 6, 3, 2, 1, 4, 5]
listB = ["a", "b", "c", "d", "e", "f", "g"]

In [5]: [i for j, i in sorted(zip(listA, listB))]
Out[5]: ['e', 'd', 'c', 'f', 'g', 'b', 'a']

for supporting C and D (removing duplicates):

sorted(list({j: i for j, i in reversed(sorted(zip(listC, listD)))}.values()))

.values() returns ListD:['a', 'c', 'e', 'f', 'g'] and .keys() returns ListC:[1, 4, 5, 6, 7]

This might help you : How to sort two lists (which reference each other) in the exact same way

For removing duplicates, you can use :

ListA = [7,6,3,2,1,4,5]
mylist = list(dict.fromkeys(ListA))

ListB = [a,b,c,d,e,f,g]
mylist = list(dict.fromkeys(ListB))

About the "removing duplicates" bit: You can start as in the other answer, but also pipe the zipped and sorted lists through a dict. In Python 3, a dict will respect the insertion order, but it will keep the last of each key, so you have to reverse the list when sorting, and then reverse back after the dict stage.

>>> listC = [1,1,4,4,5,6,7] 
>>> listD = ["a","b","c","d","e","f","g"]
>>> list(reversed(dict(sorted(zip(listC, listD), reverse=True)).items()))
[(1, 'a'), (4, 'c'), (5, 'e'), (6, 'f'), (7, 'g')]
>>> listC_new, listB_new = zip(*_)
>>> listC_new
(1, 4, 5, 6, 7)
>>> listB_new
('a', 'c', 'e', 'f', 'g')

For your first part.

listA = [7,6,3,2,1,4,5]


listB = ['a','b','c','d','e','f','g']
dict_ = {key:value for key,value in zip(listB,listA)}

listA_new = sorted(listA)
listB_new = sorted(listB,key=lambda e:dict_[e])
print(listA_new)
print(listB_new)

OUTPUT

[1, 2, 3, 4, 5, 6, 7]
['e', 'd', 'c', 'f', 'g', 'b', 'a']

For not duplicate items. Try this.

listC = [1,1,4,4,5,6,7]


listD = ['a','b','c','d','e','f','g']


listC_out = []
listD_out = []
for a,b in zip(listC,listD):
    if a not in listC_out:
        listD_out.append(b)
        listC_out.append(a)

print(listC_out)
print(listD_out)

OUTPUT

[1, 4, 5, 6, 7]
['a', 'c', 'e', 'f', 'g']