Solution 1 (official solution)
No such circular arrangement exists for $n=pq$ , where $p$ and $q$ are distinct primes. In that case, the numbers to be arranged are $p$ ; $q$ and $pq$ , and in any circular arrangement, $p$ and $q$ will be adjacent. We claim that the desired circular arrangement exists in all other cases. If $n=p^e$ where $e\ge2$ , an arbitrary circular arrangement works. Henceforth we assume that $n$ has prime factorization $p^{e_1}_{1}p^{e_2}_{2}\cdots p^{e_k}_k$ , where $p_1<p_2<\cdots<p_k$ and either $k>2$ or else $\max(e1,e2)>1$ . To construct the desired circular arrangement of $D_n:=\lbrace d:d|n\ \text{and}\ d>1\rbrace$ , start with the circular arrangement of $n,p_{1}p_{2},p_{2}p_{3}\ldots,p_{k-1}p_{k}$ as shown.

Then between $n$ and $p_{1}p_{2}$ , place (in arbitrary order) all other members of $D_n$ that have $p_1$ as their smallest prime factor. Between $p_{1}p_{2}$ and $p_{2}p_{3}$ , place all members of $D_n$ other than $p_{2}p_{3}$ that have $p_2$ as their smallest prime factor. Continue in this way, ending by placing $p_k,p^{2}_{k},\ldots,p^{e_k}_{k}$ between $p_{k-1}p_k$ and $n$ . It is easy to see that each element of $D_n$ is placed exactly one time, and any two adjacent elements have a common prime factor. Hence this arrangement has the desired property.
Note. In graph theory terms, this construction yields a Hamiltonian cycle in the graph with vertex set $D_n$ in which two vertices form an edge if the two corresponding numbers have a common prime factor. The graphs below illustrate the construction for the special cases $n=p^{2}q$ and $n=pqr$ .

Solution 2
The proof that no arrangement exists for $n=pq$ , where $p,q$ are distinct primes follows from above. Apply induction to prove all other cases are possible
Base case:
, where is a prime and is a positive integer. Any arrangement suffices , where are distinct primes. The following configuration works
$$p,pq,pr,r,qr,q,pqr$$ Inductive step: Suppose the desired arrangement exists for a composite $n$ , show the arrangement exists for $np^r$ , where $p$ is a prime relatively prime to $n$ and $r$ is a positive integer
Let $a_1,a_2,\cdots,a_m$ be the arrangement of divisors of $n$ , then $(a_i,a_{i+1})>1$ for $i=1,2,\cdots,m$ , where $a_{m+1}=a_1$ . The divisors of $np^r$ greater than 1 are of the form $$a_ip^j,p^j\qquad 1\leq i\leq m,1\leq j\leq r$$ The following sequence works $$a_1,\cdots,a_{m-1}, a_{m-1}p,\text{other divisors in arbitrary order},a_mp,a_m$$ since all other divisors are divisible by $p$ .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.