The FlT (Fermat's Little Theorem) fails for only composite numbers, and my question concerns with any pattern being followed in their failure cases - in terms of the size of the permutation set, and the number of elements in that set. If yes, then the reason is also needed. I mean by permutation set the set of permutations in a group that is separated only by rotations in the combinatorial interpretation. The combinatorial interpretation of FlT is based on a circular bead with $a$ colors, and $p$ being the number of beads, or the string size. Also, the same color strings (all same color beads) are ignored. this leads to $a$(one string for each color) subtraction from the $a^n$. This should lead to $p\mid (a^p - a)$. Also, this $a$ subtraction makes the failure cases (only for non-prime integers) being different from the not-failure cases (for prime integers), in the combinatorial interpretation.
So, if $a=2$, then there are 2 colors, $B$(Black) and $W$(White).
Coming back to the question, for example, for $a =2$ and $p=4$, there is one failure case of size two (i.e., two permutations in the set) :
i) $BWBW= WBWB$.
While for the case of $a =2$ and $p=6$, there are three failure cases with sizes $2,3,3$ respectively as:
i) $BWBWBW= WBWBWB$
ii) $BBWBBW=WBBWBB=BWBBWB$
iii) $WWBWWB=BWWBWW=WBWWBW$
I can see that $6$ factors into two factors $2$ and $3$, but that does not help.
Addendum The other or non-failure permutations for the case of $a=2$ and $p=4$ are : $$BWWW=WBWW=WWBW=WWWB \\ BBWW=WBBW=WWBB=BWWB \\ BBBW=WBBB=BWBB=BBWB,$$
The $=$ sign equates the equivalent permutations separated by a rotation only.
Similarly, there are $54$ non-failure permutations in group of $6(=p)$ are there for $a=2$ and $p=6$. The $8$ failure cases are specified above.
