Question
Let $1\lhd G_1 \lhd \ldots \lhd G_n=G$ be a composition series of the group $G$. If for every $i\not= j$ the quotients $G_{i+1}/G_i$ and $G_{j+1}/G_j$ are non isomorphic, then show that every two normal subgroups of $G$ are non isomorphic.
Attempt
If $H,K$ are two normal subgroups of $G$, then the series $1\lhd G_1\cap H \lhd \ldots \lhd G_n\cap H=H$ and $1\lhd G_1\cap K \lhd \ldots \lhd G_n\cap K=K$ are two composition series of $H,K$. I can not see how to combine them in order to show that $H\not\cong K$.
Definition: a composition series of a group $G$ is a series $1\lhd G_1\lhd\ldots \lhd G_n=G$ such that each quotient group $G_{i+1}/G_i$ is simple.
The idea behind the solution is essentially contained in the hint by @Derek Holt, which I will just expand.
Assume by contradiction that $H \cong K$, with $H, K \lhd G$ and $H \neq K$. Let $\phi: K \rightarrow H$ be the group isomorphism and set $N:=\phi(K \cap H) \lhd H$. Then $HK \lhd G$ and $$HK/H \cong K/(K \cap H) \cong H/N$$
Now consider the subnormal series $$1 \lhd N \lhd H \lhd HK \lhd G$$ and expand it to a composition series by splitting in simple factors $N$, $G/HK$ and $HK/H$ (thanks to the previous isomorphism, the last operation gives also a composition series for $H/N$). By the Jordan-Holder Theorem, that composition series will have the same quotients as $1 \lhd G_1 \lhd \ldots \lhd G_n=G$, up to isomorphisms and permutations. In particular, two different quotients of the series can never be isomorphic. But we know that each factor of $HK/H$ is isomorphic to the corresponding factor of $H/N$. So the only possibility is that both these quotients are trivial, that is $$HK/H=\{1\} \Rightarrow HK=H \Rightarrow K \lhd H$$ Since $H$ admits a composition series (as $G$ does), the isomorphism $H \cong K$ forces the identity $K=H$ (otherwise we would find two composition series for $H$ which differ for the factors contained in $H/K$, contradicting Jordan-Holder), contradiction!