 Solution

LetBoth have splitting field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. On the roots of $f$, the Galois group has two orbits of size two and embeds asThe roots of $g$, namely $\pm \sqrt{2}\pm \sqrt{3}$, form one orbit and give the regular Klein-four subgroupThe subgroups are not conjugate because their orbit decompositions differ.