More generally: given any ordered sequence of $\cong$s and $\not\cong$s, we can realize a pair of composition series whose terms are related by this sequence. In particular, if $s$ denotes the number of switches from $\cong$ to $\not\cong$ in the sequence, and $\ell$ the length of the sequence, then for any prime $p$ there are pairs of composition series of $C(p^\ell)\oplus C(p^s)$ for which this holds.
Let $f:\{1,2,\cdots,\ell\}\to \{\cong,\not\cong\}$ represent an ordered sequence of $\cong$s and $\not\cong$s, with the condition that $f(1),f(\ell)$ are both $\cong$. Let $s$ be the number of 'breaks,' or indices $1\le i< \ell$ such that $f(i)$ is the $\cong$ symbol and $f(i+1)$ is the $\not\cong$ symbol. Given a prime $p$, define $G=C(p^{\ell-s})\oplus C(p^s)$ to be a direct sum of cyclic groups, which will be a finite abelian $p$-group. We construct a pair of series $H_i$ and $K_i$ for $G$ such that $H_i\,f(i)\,K_i$ for each $1\le i\le\ell$ through a recursive alternating process.
To wit: set $H_1=C(p)\oplus C(1)=K_1$. For each $1,
if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\cong \\ f(i+1)\text{ is }\cong\\H_i,K_i=C(p^\alpha)\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\end{cases}$;
if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\cong \\ f(i+1)\text{ is }\not\cong\\H_i,K_i=C(p^\alpha)\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\end{cases}$;
if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\not\cong \\ f(i+1)\text{ is }\not\cong\\H_i=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_i=C(p^{\alpha+1})\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+2})\oplus C(p^\beta)\end{cases}$;
if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\not\cong \\ f(i+1)\text{ is }\cong\\H_i=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_i=C(p^{\alpha+1})\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1})\end{cases}$.
In other words, we create two paths in $\Bbb N^2$ starting at $(1,0)$: $\cong\to\cong$ and $\not\cong\to\not\cong$ correspond to adding the vector $(1,0)$ to both paths; $\cong\to\not\cong$ corresponds to adding $(0,1)$ and $(1,0)$ to the paths resp. and $\not\cong\to\cong$ corresponds to adding $(1,0)$ and $(1,0)$ to the paths resp.
As an example, for $p=2$ and $(f(i))_{i=1}^4=(\cong,\not\cong,\not\cong,\cong)$, we have for $C_8\oplus C_2$,
$\begin{array}{|c|c|c|c|c|}\hline i & 1 & 2 & 3 & 4 \\ \hline H_i & C_2\oplus C_1 & C_2\oplus C_2 & C_4\oplus C_2 & C_8\oplus C_2 \\ \hline K_i & C_2\oplus C_1 & C_4\oplus C_1 & C_8\oplus C_1 & C_8\oplus C_2 \\ \hline \end{array}.$
It is then easy to see, setting $H_0=C_1=K_0$ for convenience, that $\frac{H_4}{H_3}\cong\frac{H_3}{H_2}\cong\frac{H_2}{H_1}\cong \frac{H_1}{H_0}\cong C_2\cong\frac{K_1}{K_0}\cong\frac{K_2}{K_1}\cong\frac{K_3}{K_2}\cong\frac{K_4}{K_3}, \\ H_1\cong K_1,\quad H_2\not\cong K_2,\quad H_3\not\cong K_3,\quad H_4\cong K_4.$
Note that, for every positive integer and divisor, $d\mid n$, we view $C_d$ as a subgroup of $C_n$.
This extends Ted's examples so that every question is answered: every sequence can be realized, no subset of terms can be set isomorphic in order to force the other terms to be isomorphic, and this applies to chief series (since all subgroups of abelian groups are normal) which are examples of normal, subnormal, and composition series as well. All with finite abelian $p$-groups of two factors.