Action of finite group on a set of subgroups is transitive Let $G$ be a finite non-trivial group and $X$ be a non-empty set of all subgroups of $G$ which satisfy the following:
(1) Any two elements of $X$ are either disjoint (i.e., intersection is trivial) or identical.
(2) For all $H\in X$, we have $N_G(H)=H$.
(3) $G$ acts on $X$  by conjugation.
Show that all elements of $X$ are conjugates.
My attempt: We need to show that the action of $G$ on $X$ is transitive. If $G$ is Abelian, then all its subgroups are normal and so $X=\{G\}$, in which case we're done. So assume that $G$ is not Abelian. Then $|G|\geq 6$. If $|G|=6$, then $G=S_3$. In this case, $X$ becomes the set of all Sylow-2 subgroups of $G$ and so we're done by Sylow's theorem.
By induction, assume that the result holds for all groups with order less that $|G|$ and fix $H\in X$. Consider the action of $H$ on $X$. Since $|H|<|G|$, this action is transitive by induction hypothesis. In particular, all elements of $X$ are conjugates.

Is my "proof" correct? If yes, are there any alternate ways to prove the result? If not, how do I proceed?

Edit: As @lulu pointed out, my solution is incorrect. Other attempts I have made are that I have shown that ${\rm Stab}_G(H)=H$ for all $H\in X$ and so by orbit-stabilizer theorem, $1<|{\rm Orb}(H)|=[G:H]$ for all $H\in X$.

I don't know how to proceed from here.

 A: In the assumptions of the problem, Burnside's (counting) lemma yields:
\begin{alignat}{1}
\#\text{ of orbits} &= \frac{1}{|G|}\sum_{H\in X}|H| \\
&\le \frac{1}{|G|}(|G|-1+|X|) \\
&=1+\frac{|X|-1}{|G|} \\
&<2
\end{alignat}
and hence $\#\text{ of orbits}=1$.
A: If $H \in X$, then the $G$-conjugates of $H$ form a TI set, so there are $|G:H|$ such conjugates and the union of their non-identity elements has size $|G|(|H|-1)/|H| = |G|- |G:H|$, and hence there are only $|G:H|-1$ non-identity elements that do not lie in a conjugate of $H$.
So there are not enough elements left for another subgroup $H_2 \in X$ that is not conjugate to $H$, and so $X$ consists of the $G$-conjugates of $H$.
Equivalently, a finite group has at most one faithful action that is a  Frobenius group.
A: Let me elaborate Derek's argument: If $H_2\in X$ is not conjugate to $H$, then by the same argument, there are $|G|-|G:H_2|$ non-trivial elements in the conjugates of $H_2$. Since they do not lie in conjugates of $H$, we obtain
$$|G:H_2|\le|G:H_2|(|H_2|-1)=|G|-|G:H_2|\le|G:H|-1.$$
Interchanging $H$ and $H_2$ yields a contradiction.
