Let $n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$ with the $p_i$ distinct primes and $ \alpha_i\in \Bbb N$. Just to check if I'm correct, is it true that $k!$ is the number of order-isomorphisms of the form $f:(D_n,|)\to (D_n,|)$?
$D_n$ is the lattice of all positive divisors of $n$.
Your conjecture is only true when all exponents are equal, that is, when $n$ is of the form $\big(p_1\cdots p_k\big)^{\alpha}$ for some $\alpha\geq 1$.
An order isomorphism $\phi:D_n\to D_n$ induces a permutation on the immediate successors of the bottom element $1$ (which necessarily is a fixed point), i.e. there exists a permutation $\sigma\in\mathfrak{S}_k$ such that $\phi(p_i)=p_j$ where $j=\sigma(i)$. Actually, for all $1\leq a\leq\alpha_i$, $$\phi(p_i^a)=p_j^a$$ (again, $j=\sigma(i)$). This can be seen inductively on $a$ : the prime powers $p^a$ have the unique property that they are the immediate successors of exactly other element (that is $p^{a-1}$), whereas composite numbers are always immediate successors to $d$ elements, where $d\geq 2$ is the number of prime factors of said number. This in turn completely determines $\phi$, since it will preserve joins.
It follows that $\sigma$ cannot be totally random, it must preserve the partition $P_1\sqcup\cdots\sqcup P_r$ of $\lbrace 1,\dots, k\rbrace$ defined by the equivalence relation $i\sim j$ iff $\alpha_i=\alpha_j$.
This is easily seen to be the only restriction on $\sigma$, so that the group of automorphisms is isomorphic to a product of symmetric groups $$\mathrm{Aut}(D_n)\simeq\prod_{i=1}^r\,\mathfrak{S}_{c_i}$$ where $c_i=\#P_i$ and $r$ is (by definition) the number of different exponents $\alpha_i$ in the prime decomposition of $n$.