Let an integer N be the number of digits imprinted on a bracelet, which can come in two values, 1 and 0.
You can produce a binary number by writing down the 1's and 0's on the bracelet from left to right, starting at an arbitrary point, until you wrap around completely. Since you started at an arbitrary point, a bracelet with N digits can produce up to N different binary numbers.
For example, the numbers A=1110 and B=0111 come from the same bracelet, since B is produced by shifting all digits of A to the right and looping the right-most digit to the left.
For the following question, let us ignore bracelets that repeat before you loop them N times, for example 0101, 1111, 0000000, or 01100110, and trivial examples (like 0, 1):
Question: If $a_1$, $a_2$, ..., $a_N$ are all the binary numbers associated with permutations of a unique bracelet with N elements, then they do not all share a common factor (at least two are co-prime). EDIT: For N>5
Giving a quick example, the bracelet 000101 can also come in the forms of 100010, 010001, 101000, 010100 and 001010. This translates to the numbers 5, 34, 17, 40, 20, 10. They do not all share a common factor (except for 1).
Bonus question: Does every unique N-bracelet contain a permutation that is a prime number? (not counting bracelets that repeat before looping N times)
As it turns out, it is easy to generate counterexamples, since divisibility by $3$ is preserved by this operation.
Note that a simple divisibility test for division by $3$ is to take the alternating sum of the binary digits, since $2\equiv -1\pmod 3$. Thus for instance $$1001000010000001_2$$
Is divisible by $3$ since there are exactly two $1's$ in each of the even and the odd numbered slots.
It is not difficult to see that every cyclic permutation of this shares this property, hence $3$ is a common factor. In particular, none of these are prime. It is easy to generalize this to arbitrarily long numbers.
Note: the same principle works in different bases. In base $b$, just use a prime which divides $b\pm 1$. Hence, in base $10$, you can use $3$ or $11$. To make the question interesting, you need to ignore those prime divisors.