Is there a finite group $G$ and a divisor $d$ of $|G|$ so that $G$ contains exactly two subgroups of order $d$?

The motivation for this question is an old qual problem (see http://www.math.wisc.edu/~passman/qualjan01.pdf, problem 1). If $G$ is such a group and $A$ and $B$ are the two subgroups of order $d$, then it is easy to see that $A \unlhd \langle A, B \rangle$ and $B \unlhd \langle A, B \rangle$ and so $$ \langle A,B \rangle / (A \cap B) \cong A/(A \cap B) \times B/(A \cap B) $$ is another group (with possibly smaller order) with this property. Assume therefore that $G \cong A \times B$ and the only two subgroups of $G$ with order $d$ are $A \times 1$ and $1 \times B$. Since $(\mathbb{Z}/p\mathbb{Z})^{2}$ has $p+1$ subgroups of order $p$, we see that $d$ is not prime.

Moreover, if $A$ has a subgroup $C$ of index $p^{k}$, then $B$ has a subgroup of order $p^{k}$, then $C \times D$ is another subgroup of $G$ that also has order $d$. This implies that neither $A$ nor $B$ can have any subgroups of prime power index, and this implies that they are both perfect.

I searched Magma's database of perfect groups for pairs $(A,B)$ with the same order. For each pair $(A,B)$, there is a subgroup of $A \times B$ of the form $C \times D$, where $C$ is a non-trivial subgroup of $A$ and $D$ is a non-trivial subgroup of $B$.

Note: I remember asking Isaacs this question. His recollection was that this was once discussed on the Group Pub Forum, and that the answer was yes, but he didn't recall details of the construction.