Let the rearranged numbers be $a_{1}, \ldots, a_{25}$. The number of pairs $(n, m)$ with $n \mid m$ must equal the number of pairs with $a_{n} \mid a_{m}$, but since each pair of the former type is also of the latter type, the converse must be true as well. Thus, $n \mid m$ if and only if $a_{n} \mid a_{m}$. Now for each $n=1,2, \ldots, 6$, the number of values divisible by $n$ uniquely determines $n$, so $n=a_{n}$. Similarly, 7,8 must either be kept fixed by the rearrangement or interchanged, because they are the only values that divide exactly 2 other numbers in the sequence; since 7 is prime and 8 is not, we conclude they are kept fixed. Then we can easily check by induction that $n=a_{n}$ for all larger composite numbers $n \leq 25$ (by using $m=a_{m}$ for all proper factors $m$ of $n$ ) and $n=11$ (because it is the only prime that divides exactly 1 other number). So we have only the primes $n=13,17,19,23$ left to rearrange, and it is easily seen that these can be permuted arbitrarily, leaving 4 ! possible orderings altogether.