Let $f \in \Bbb Q[X]$ be an irreducible polynomial with complex roots $a_1, .., a_n$ and splitting field $K$. The action of $G:=Gal(K/\Bbb Q)$ on the roots gives an embedding $\rho : G \to S_n$.
If $\rho(g)$ and $\rho(h)$ are conjugate in $S_n$, does it follow that the subgroups $\langle g \rangle, \langle h \rangle$ are conjugate in $G$ (for all $g,h \in G$) ? I'm not sure if this is true.
It does not follow. Example: $\newcommand{\QQ}{\Bbb{Q}}\QQ(i,\sqrt{2})$. This has Galois group $C_2\times C_2$ generated by complex conjugation and $\sqrt{2}\mapsto -\sqrt{2}$. This acts on the four elements $i+\sqrt{2}$, $i-\sqrt{2}$, $-i+\sqrt{2}$, and $-i-\sqrt{2}$ by always simultaneously swapping two pairs of elements (for nonidentity elements of the Galois group). Thus it embeds as the normal Klein 4 subgroup of $S_4$. In particular, the nonidentity elements of the subgroup have the same cycle structure in $S_4$, and are thus conjugate in $S_4$. However the Galois group itself is abelian, so none of the subgroups are conjugate to each other in $G$.