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The Task
In this challenge, your task is to write a program, which takes in no input, with as many anagrams that are quine of themselves as possible.
Your score will be the number of anagrams of your program's source code that are valid quines divided by the total number of anagrams possible, i.e. the percentage of the anagrams that are valid quines.
Standard Loopholes and rules of standard quines apply.
Note: Your program must have at least 3 characters (not 3 bytes).
Input
Each anagram (or permutation) of your program which is quine of itself (i.e. the anagrams which you are including in your score) must not take any input. If your language requires input as a necessity, then you can assume that your program will be given a String consisting of the lowercase letter A. However, you must not make any use of the input in any way.
Output
n number of distinct anagrams of your program's source-code must be valid quines, where n represents the number of those anagrams that you are including in your score, i.e.
Those quine-anagarams can output in any way except writing the output to a variable. Writing to file, console, screen etc. is allowed. Function return is allowed as well.
Scoring Example
Suppose your program's source code is code. And,
codeoutputscode.coedoutputscoed.cdoeoutputscdoe.cdeooutputscdeo.cedooutputscedo.ceodoutputsceod.ocdeoutputsocde.ocedoutputsoced.odceoutputsodce.odecdoes not outputodecor produces error.oedcdoes not outputoedcor produces error.oecddoes not outputoecdor produces error.docedoes not outputdoceor produces error.doecdoes not outputdoecor produces error.dcoedoes not outputdcoeor produces error.dceodoes not outputdecoor produces error.decodoes not outputdecoor produces error.deocdoes not outputdeocor produces error.eodcdoes not outputeodcor produces error.eocddoes not outputeocdor produces error.edocdoes not outputedocor produces error.edcodoes not outputedcoor produces error.ecdodoes not outputecdoor produces error.ecoddoes not outputecodor produces error.
The score of this solution will be
Winning Criterion
The solution with the highest score wins! In case of a tie, the answer with higher char-count (not byte-count) wins (hence the code-bowling)! If a tie still persists, then the solution which was posted earlier wins!


I believe this is a duplicate of https://codegolf.stackexchange.com/questions/121522/rotation-safe-quine
– Neil A. – 2017-05-21T08:01:14.213@NeilA. I'm a lot more confident that a rotation-safe quine can actually be written in other languages than Unary (although not in many languages and it will still be hard). – Martin Ender – 2017-05-21T08:12:31.243
It's probably too late to change the specs now, but you should probably have specified that the original program doesn't count, because otherwise this essentially seems to be a duplicate of our vanilla quine question. – SuperJedi224 – 2017-05-23T01:50:16.743