Suppose $G$ is a finite group, $A,B$ are subgroups of $G$ such that they are the only subgroups of order $n \in \mathbb{Z}^+$.
Suppose $X$ is the set of subgroups of $G$ such that $G$ acts on $X$ by conjugation of subgroups, i.e. $g\cdot K = \{gkg^{−1} : k \in K\}$, if $g \in G$ and $K \in X$.
Suppose that $|G|$ is not even. Prove that $A$ and $B$ are both normal in $G$.
I have worked out that if $K \in X$ then, if $A \in {\rm Orb}_G(K)$ then $|A| = |K|$.
I want to use the fact that $|{\rm Orb}_G(K)| = {|G|\over|{\rm Stab}_G(K)|}$ from the Orbit-Stabiliser Theorem but not sure where to go with it.
As you say, $A \in Orb_G(K)$ implies $|A| = |K|$. Therefore $|Orb_G(A)|$ is either equal to $1$ or $2$, since $B$ is the only other subgroup which could be in the orbit of $A$. If $|Orb_G(A)| = 1$ then $gAg^{-1} = A$ for all $g \in G$ so $A$ is normal, and $B$ must be too since if $B$ is not in the orbit of $A$ then $A$ is not in the orbit of $B$. However, if $|G|$ is odd then $|Orb_G(A)|$ cannot be equal to $2$ because that would imply $|Stab_G(A)| = \frac{|G|}{2}$.