Since I haven't had anyone reply, either it is not an easy problem to solve or I didn't explain it adequately.

Since the software shifted the top line of the mod 5 matrix, (the top line is the exponent line), let me provide a mod 7 matrix for clarity. I have to eliminate the subscripts since the software doesn't allow it's use for replies. I would also like to correct an error that read "the next to the next column" should just read "the next to the last column".

X 2 3 4 5 6 7

1 1 1 1 1 1 1

2 4 1 2 4 1 2

3 2 6 4 5 1 3

4 2 1 4 2 1 4

5 4 6 2 3 1 5

6 1 6 1 6 1 6

7 0 0 0 0 0 0

Note that n-2 = 5, and the column under the exponent 5 (and all exponents congruent to 5 mod 7) has all unique numbers. Note also that the exponent 6 affirms Fermat's Little Theorem, and the exponent 7 affirms that a^n == a mod n. Most important for my conjecture is that the third from the last column contains all unique number congruences, without duplication.