OK, continuing the research I want to tell some facts:
1) Before permutating (rearranging) columns with key, free squares are filed with letter X, for example:
K E Y K E Y
1 F A G 1 F A G
2 V D F <=> 2 V D F
3 F A A 3 F A A
4 F 4 F X X
This means the final encrypted text length will be divisible by key length.
In this example length is 12 and key has 3 letters.
Now we our ciphertext's length is 48
It means we have several possible key lengths, as 48 is divisible by:
1, 2, - Hardly possible
3, 4, 6, 8, 12, - Possible
16, 24, 48 - Hardly possible
Of which most probably key length is 3, 4, 6, 8, 12.
Maybe now we should find words with this lengths:
JohnsRevenge - 12
John - 4
Kslkgh ?- 6
ADFGVX - 6
2) Empty boxes are filled with X. It means that last row can have more X-es, because count of X on last row doesn't change even after permutation.
Also probablity of X* pair the least, because usually all unused letters/numbers are in X row (in non-permutated ciphertext).
A D F G V X
V B N 0 1 2 3
X-> 4 5 6 7 8 9 - symbols in this row are used rarely, so X* is rare
Back to the game. When trying all possible key lengthes (write cipher VDFVX... upside down, left to right, in 3, 4, 6, 8, 12 columns) the more X-es we have on the last row the more possibility is that we have found the correct key length. (It's about possibility, not 100% correct)
3) Not a fact but interesting:
@Kslkgh, I think I know the cipher and I think I know the key, but I can't for the life of me figure out what the keyword is...am I on the right track?
@BaileyM Do you mean that you know what the ->keysquare<- is. If you do, you are definately on the right track! – Kslkgh
Seems like he the OP reacted more on keysquare (and doesn't care about permutation key?). So probably key is obvious/easy to find.
The key is hidden in a ->method<- I've used on various occasions – Kslkgh again makes me think the key is method name- ADFGVX (also it's length is equal to 6, and 48 is divisible by 6).
TO BE CONTINUED...