Are there groups with conjugacy classes as large as the divisors of perfect numbers? Are there groups, where the conjugacy classes have elements according to the divisors of perfect numbers larger than $6$?
I already got the first example, so therefore $6$ is excluded: $S_3$.
The corresponding GroupProps page on groups of order $28$ is still empty, but OEIS/A000001 says that there are $4$ groups having $28$ elements. 
Any idea?
 A: Suppose $|G|=n$, and that $n$ is a perfect number, with $G$ having a conjugacy class of every possible size. Let's show $n=6$.
First, let $p$ be the smallest prime dividing $n$, and let $x\in G$ be an element with conjugacy class size $n/p$. Then $C_G(x)$ has size $p$, so $x$ is an element of order $p$ that is self-centralizing. Since $N_G(\langle x\rangle)/C_G(\langle x\rangle)$ has order dividing $p-1$, the minimality of $p$ implies $N_G(\langle x\rangle)=C_G(\langle x\rangle)$. Thus $G$ has a normal $p$-complement, so that $G=H\rtimes \langle x\rangle$, with $|H|=n/p$.
Now let $y\in H$ have a conjugacy class size of $p$.  Then the centralizer of $y$ has order $|H|$, so $y\in Z(H)$. This means that if $|y|=k$, then every element in $H$ has centralizer at least of order $k$, so conjugacy class size dividing $n/k$.  Since we want conjugacy classes of every possible size, this is only possible if $n/k=p$, or $n=kp$. In particular, $H$ is cyclic with generator $y$.
But then the only possible conjugacy classes are those in $H$ of size $p$, and those outside $H$ of size $n/p=k$.  Thus $k$ must be a prime $q$, and $n=pq$.  By the assumption that $n$ is perfect, $pq=p+q+1$. Since $p<q$, working mod $q$ shows
$$ p+1=q.$$
Thus $p=2$, $q=3$, and $n=6$.
