Although the two strings have different lengths and the characters are in a different order, they still have the same count of each alpha character. In your examples " abcdkazbmopdkubmgt " and "abckdazmbopkdumbgt" both have 2xAs, 3xBs, 1xC, 2xDs, 1xG, 2xKs, 2xMs, 1xO, 1xP, 1xT, 1xU and 1xZ (and no other alpha).
Here's a function that will count each alpha in your string. Note, I'm a novice so there's probably a more efficient coding method...
CREATE FUNCTION [dbo].[FN_CharID]
declare @position as int
declare @ComparisonString as char(26) = 'abcdefghijklmnopqrstuvwxyz'
declare @CurrentLetter as char(1)
declare @cnt as int
declare @tmpresult as char(26)
set @position = 1
set @tmpresult = '00000000000000000000000000'
while @position < 27
set @cnt = 0
set @CurrentLetter = substring(@ComparisonString,@position,1)
WHILE charindex(@CurrentLetter,@StringtoAssess) > 0
set @cnt = @cnt + 1
set @StringtoAssess = stuff(@StringtoAssess, charindex(@CurrentLetter, @StringtoAssess), 1, '')
set @tmpresult = stuff(@tmpresult, @position, 1, @cnt)
set @position = @position + 1
set @StringtoAssess = @tmpresult
Running the function across your strings both return "23120010002020110001100001". I hope this is useful.