Why does sort() with key function not do anything while sorted() is working?

I have a list of integers with duplicates and I need to sort it by the number of these duplicates.

For example:

input: n = [2, 4, 1, 2]
output: n = [4, 1, 2, 2]

I wrote some code and noticed, that sort() does not change the list. But if I try to use sorted() with the same key argument, then it works just fine. What is the reason behind this?

My code:

nums = [2, 4, 1, 2]
nums.sort(key = lambda x: nums.count(x))
print(nums)

Can it be connected to sort() method using in-place algorithm?

Unlike sorted, the list.sort method sorts a list in-place, during which time the list is in an interim state where there is no integrity to its internal data structure for the other methods to read from.

Since a key function for the sort method is called during a sort, your calling the count method of the same list in the key function does not actually return the count of a given item as you expect, which you can see with a wrapper function:

def count(x):
    n = nums.count(x)
    print(n)
    return n

nums = [2, 4, 1, 2]
nums.sort(key=count)

the above outputs:

0
0
0
0

which explains why the list remains in the same order since the key for each item turns out to be the same value of 0 and since the sorting algorithm used is a stable one.

The documentation of list.sort also notes:

CPython implementation detail: While a list is being sorted, the effect of attempting to mutate, or even inspect, the list is undefined. The C implementation of Python makes the list appear empty for the duration, and raises ValueError if it can detect that the list has been mutated during a sort.

So CPython's implementation makes the list appear empty during a sort, which explains why list.count returns 0 when called from a key function.