Solution:

Suppose the digits are $a_1, a_2, a_3, a_4, a_5, a_6$ and that $a_1a_2a_3a_4a_5a_6$ is divisible by $37$. We claim that $a_2a_3a_4a_5a_6a_1$ is also divisible by $37$. Put $n = a_2a_3a_4a_5a_6$ and $m = a_1$. The original number is $10^{5}m + n$ and the derived number is $10n + m$. But $37$ divides $10^{3} - 1$ and hence also $10^{6} - 1$ and $m(10^{6} - 1)$. So it also divides $10(10^{5}m + n) - m(10^{6} - 1) = 10n + m$, which proves the claim.

Iterating, we get the original number and $5$ others:

a1a2a3a4a5a6
a2a3a4a5a6a1
a3a4a5a6a1a2
a4a5a6a1a2a3
a5a6a1a2a3a4
a6a1a2a3a4a5

Similarly, we have that $a_1a_2a_3a_4a_5a_6 - a_1a_2a_6a_4a_5a_3 = (a_6- a_3)999$, so $a_1a_2a_6a_4a_5a_3$ is also divisible by $37$. Iterating we get:

a1a2a6a4a5a3
a2a6a4a5a3a1
a6a4a5a3a1a2
a4a5a3a1a2a6
a5a3a1a2a6a4
a3a1a2a6a4a5

Similarly, we could swap the first term and the fourth to get:

a4a2a3a1a5a6
a2a3a1a5a6a4
a3a1a5a6a4a2
a1a5a6a4a2a3
a5a6a4a2a3a1
a6a4a2a3a1a5

or the second and the fifth to get:

a1a5a3a4a2a6
a5a3a4a2a6a1
a3a4a2a6a1a5
a4a2a6a1a5a3
a2a6a1a5a3a4
a6a1a5a3a4a2