Given a number field $K$, when is its Hilbert class field an abelian extension of $mathbb{Q}$?

I happened to come across this question again today. In some cases at least, the Hilbert class field $H$ of an abelian extension $K$ of $mathbf{Q}$ will have to be abelian over $mathbf{Q}$ for purely algebraic reasons.

Let $F$ be any field, $K|F$ an abelian extension of group $G=mathrm{Gal}(K|F)$ and containing a primitive $n$-th root of unity for some $n>1$, $omega:Gto(mathbf{Z}/nmathbf{Z})^times$ the cyclotomic character giving the action of $G$ on $mu_n$, and $H|K$ an abelian extension of exponent dividing $n$. Then $H=K(root nof D)$ for some subgroup $Dsubset K^times/K^{times n}$, by Kummer theory. It can be checked that $H|F$ is galoisian if and only if $D$ is $G$-stable. When such is the case, the conjugation action of $G$ on $mathrm{Gal}(H|K)$ coming from the short exact sequence
$$
1tomathrm{Gal}(H|K)tomathrm{Gal}(H|F)to Gto1
$$
is trivial if and only if $G$ acts on $D$ via $omega$. In this situation ($H=K(root nof D)$ for some subgroup $Dsubset(K^times/K^{times n})(omega)$), a sufficient condition for $H$ to be abelian over $F$, is that the order of $G$ be prime to $n$, because then $mathrm{Gal}(H|F)=mathrm{Gal}(H|K)timesmathrm{Gal}(K|F)$.

I'm sure this situation can be realised when $F=mathbf{Q}$, for example when the finite abelian extension $K$ has odd degree $[K:mathbf{Q}]$, $n=2$, the class group of $K$ has order ($1$ or) $2$, and $H$ is the Hilbert class field of $K$. In this case the extension $H|mathbf{Q}$ will be necessarily abelian.