Question: Let $G$ be a group of order $n^2$, and $H$ a subgroup of order $n$. prove that $H$ contains a non-trivial normal subgroup of $G$.
Remarks:
-It's equivalent to prove that $\cap_{x\in G} H^x$ is a non-trivial subgroup of $G$. because the intersection is the maximal normal subgroup of $G$ contained in $H$.
-The intersection of any two conjugates of $H$ is non-trivial. because if for some $x,y\in G$, $H^x\cap H^y=\{e\}$ then $$ |H^xH^y|=\frac{|H^x||H^y|}{|H^x\cap H^y|}=n^2 $$ hence $G=(H^x)(H^x)^{x^{-1}y}$, which is impossible.
Let $H$ be a subgroup of $G$ such that $|H|=n, |G|=n^2$. Then $H$ has $n$ cosets inside $G$ and $G$ acts on these cosets on the left. Thus, we get a map $\rho: G \to S_n$. The kernel of $\rho$ is a normal subgroup, and any element $g \in \mathrm{Ker}(\rho)$ must satisfy $gH=H$. So $\mathrm{Ker}(\rho) \subset H$.
From the above discussion, we see that a negative answer to your question is the same as the above map having trivial kernel. In other words, we can construct a counter-example by finding a transitive subgroup of size $n^2$ in $S_n$. The smallest possible $n$ which could yield a counter-example is $n=6$ and the comments give a counter-example in this case.