Regular expression that matches only the letters in a given word

I am trying to create a regex expression which will allow the test string to match with the following conditions:

1. The test string can be any length and contain any letters.
2. ALL characters in the regex must be matched.
3. All letters in regex must be matched ONLY the number of times they are repeated in the regex expression.
4. If a letter is repeated more than once in the regex it can be consecutive in the test string or separated by other letters.
5. The letters in the test string don't have to be in the same order as the regex.

So if:

TEST STRING = polymmorphiiic

the following regex would match

ppy
ooyc
ciii

whereas the following would fail:

pmy
iy
yrz

Your help appreciated as I can get all the above satisfied.

I have tried breaking down all the requirements and creating regex string for each but I can't get over string that will cover all.

You can use re.findall() to get the matching characters out of the string. A comparison between the sorted result and the sorted pattern will handle the differences of order and character counts.

import re

testString = "polymmorphiiic"

for p in ("ppy", "ooyc", "ciii", "pmy", "iy", "yrz"):
    print(p,"matches:",sorted(p) ==  sorted(re.findall(f"[{p}]",testString)))

ppy matches: True
ooyc matches: True
ciii matches: True
pmy matches: False
iy matches: False
yrz matches: False

Note that you don't need a regular expression to implement this process, a set is sufficient to filter on the matching characters:

for p in ("ppy", "ooyc", "ciii", "pmy", "iy", "yrz"):
    matches = set(p).intersection
    print(p,"matches:",sorted(p) == sorted(filter(matches,testString)))

Even a sequential search would work:

for p in ("ppy", "ooyc", "ciii", "pmy", "iy", "yrz"):
    print(p,"matches:",sorted(p) == sorted(filter(p.__contains__,testString)))