Solution:
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.