$\textbf{Problem.}$ Let $G$ be a solvable group of order $mn$, with $m,n>1$ and $(m,n)=1$. Let $H,\bar{H},K,\bar{K}$ be subgroups of $G$ such that $|H|=|\bar{H}|=m$ and $|K|=|\bar{K}|=n$. Show that there is $g \in G$ such that $\bar{H}=H^g$ and $\bar{K}=K^g$.
I tried to use Hall's theorem, to guarantee the existence of subgroups of order $m$ and $n$, and any others are conjugated, but I couldn't show why it's possible to get the same element $g$ in this case.
Outline proof:
(i) Since $H$ and $\bar{H}$ are conjugate, we can assume that $H = \bar{H}$, and we need to find $g \in G$ with $H^g = H$ and $K^g = \bar{K}$.
(ii) Since $|G|=mn$, we have $G = KH$. Let $g' \in G$ with $K^{g'} = \bar{K}$. Then we can write $g' = kh$ with $k \in K$, $h \in H$, and now $g = h$ works.
To see that you can assume that $H = \bar{H}$, start by choosing $x \in G$ with $H^x = \bar{H}$. Now apply the proof above but with$\bar{K}$ replaced by $\bar{K}^{x^{-1}}$, and find $g \in G$ with $H^g=H$ and $K^g = \bar{K}^{x^{-1}}$. Now $H^{gx}=\bar{H}$ and $K^{gx} = \bar{K}$.