 Solution

A composition series is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$ with simple nontrivial factors. Every finite group has one, by recursively choosing a maximal proper normal subgroup. The Jordan–Hölder theorem states that any two such series have the same length and the same multiset of simple factors up to isomorphism.

For a proof induct on $|G|$. Let $M,N$ be the penultimate terms in two series. If $M=N$, apply induction there. Otherwise their maximal normality gives $MN=G$. Set $D=M\cap N$. The second isomorphism theorem givesso these quotients are simple. Choose a composition series of $D$ and extend it once to $M$ and once to $N$. Induction in the smaller groups $M,N$ compares these with the respective original prefixes. Appending $G/M$ or $G/N$ gives in both cases the factors of $D$ together with $G/M,G/N$. This completes the proof, including the case $D=1$.

Exactly one: take $S_5$. Its only nontrivial proper normal subgroup is $A_5$. Indeed a normal $N$ has $N\cap A_5$ either $1$ or $A_5$ by the supplied simplicity. In the first case $N$ injects into $S_5/A_5\cong C_2$, so a nontrivial $N$ would be a normal subgroup of order two and hence central. But $S_5$ has trivial center: conjugating every transposition trivially forces preservation of each unordered pair, and therefore of every point. Thus the unique series is $1◃A_5◃S_5$.

Exactly two: take $A_5\times C_2$. A normal subgroup containing $A_5\times 1$ is that subgroup or the whole product. For a normal subgroup $N$ meeting $A_5\times 1$ trivially, commutators of its elements with $A_5\times 1$ vanish; its first coordinates therefore lie in $Z(A_5)=1$. Hence $N\subseteq 1\times C_2$. The two nontrivial proper normal subgroups are exactly the two factors, each simple, so the two series pass respectively through $A_5\times 1$ and $1\times C_2$.

More than two: take $C_{120}$. A cyclic group has exactly one subgroup of each divisor order, and composition steps have prime indices. The orders of those indices are the distinct permutations of $2,2,2,3,5$, givingAll three example groups have order $120$.