Originally Posted by adfgvx
Here is what I think:
At first glance I thought that this was an ADFGVX cipher with the letters substituted.
For those who don't know what this is, its basically a cipher in which we have a 6x6 grid, 36 boxes, filled in with the 26 letters of the alphabet as well as the 10 numbers. Bordering the top of the grid are the letters ADFGVX as shown below:
__A D F G V X
To encrypt, one finds their plaintext letter in the grid, and looks horizontally to see which row, and vertically for which column, and uses the corresponding letters as ciphertext (digraphs).
The reasoning that this may be that kind of cipher is because it has digraphs, and the cipher seems to be made with repetitive letters. However, an ADFGVX consists of 6 distinct letters, and as someone mentioned before, there are actually 10 of them: O,N,T,P,Y,U,V,S,R,Q. While it is a bit far-fetched, may I propose that this cipher is built on a similar principle?
Lets say we have 5x5 grid, on which the outside edges are bordered by the 10 distinct letters. There are 26 letters in the alphabet, and so two letters must share a box in the grid, maybe I/J as with a playfair or probably W/X. The reason for this is because it's interesting how 9 letters, N through V are in alphabetical order, whilst Y is on its lonesome. Perhaps the gap of WX hints that those two letters share a box.
Now moving on the grid itself. The key to the grid may lie in the selection of letters represented in the cipher, as in, how it starts out with N leading on to V and then Y. Perhaps it is filled out in alphabetical order with those letters omitted, and then added on to the end. Another idea is that because N is the 14th letter in the alphabet, the key might be the word FOURTEEN, or THIRTEEN, if A is 0. Lastly, the common occurrence of N suggests that the row or column it dictates has the five vowels in it, A, E, I, O, and U.
As to why there are four letters and six letters grouped together, it may just be a decoy, and that we can split them up into pairs and look at them separately.
That's just my input, and what I think. If anybody thinks I'm onto something or I've made a grievous error, than please post to let me know.