Subgroups of finite abelian groups and automorphisms Let $A$ a finite abelian group, and $B$, $C$ finite subgroups of $A$. 
If $B$, $C$ are isomorphic, and their quotient groups $A/B$, and $A/C$ are isomorphic, it may still happen that there does not exist an automorphism of $A$ that takes $B$ to $C$. See a question on MSE at this link, and the solution at this page. Note that the implication holds if $B$ (or $A/B$) has prime order (not that hard to prove).
I wonder if the following is true, or if anyone has a counterexample:
Let $A$ be a finite abelian group, $B$, $C$ isomorphic subgroups. Assume that for every(many)  characteristic subgroup $B'$ of $B$, and corresponding $C'$ of $C$, the quotient groups $A/B'$, $A/C'$ are isomorphic. Then there exists an automorphism of $A$ that takes $B$ to $C$. 
 A: There is a counterexample. Given any prime $p$, let
\begin{align}
A 
&= \begin{pmatrix} \mathbb Z/p^5 & \mathbb Z/p^2 \\ \mathbb Z/p & \mathbb Z/p^4  \end{pmatrix} \\
&:= \left\{ \begin{pmatrix}x & y \\ z & w\end{pmatrix} \middle|\, x \in \mathbb Z/p^5, y \in \mathbb Z/p^2, z \in \mathbb Z/p, w \in \mathbb Z/p^4 \right\} \\
&\cong \mathbb Z/p^5 \times \mathbb Z/p^2 \times \mathbb Z/p \times \mathbb Z/p^4,
\end{align}
let $B$ be generated by 
$$b_1 = \begin{pmatrix} p^3 & 0 \\ 1 & 0\end{pmatrix}\quad \text{and}\quad b_2 = \begin{pmatrix}0 & p \\ 0 & p^2 \end{pmatrix},$$
and let $C$ be generated by
$$\begin{pmatrix}p^3 & p \\ 0 & 0 \end{pmatrix} \quad \text{and}\quad \begin{pmatrix}0 & 0 \\ 1 & p^2 \end{pmatrix}.$$
Thinking of $A$ columnwise as the product
$$A = A_1 \times A_2 = (\mathbb Z/p^5 \times \mathbb Z/p) \times ( \mathbb Z/p^2 \times \mathbb Z/p^4),$$
we have $B = \langle b_1 \rangle \times \langle b_2 \rangle \cong (\mathbb Z/p^2) \times (\mathbb Z/p^2),$ so the only characteristic subgroup of $B$ other than $B$ itself and the trivial group is 
$$pB = \langle p b_1 \rangle \times \langle p b_2 \rangle \cong \mathbb Z/p \times \mathbb Z/p.$$
It's straightforward to check that
$$A/B = A_1/\langle b_1 \rangle \times A_2/\langle b_2 \rangle \cong (\mathbb Z/p^4) \times (\mathbb Z/p \times \mathbb Z/p^3)$$
Similarly, thinking of $A$ rowwise as
$$A = (\mathbb Z/p^5 \times \mathbb Z/p^2) \times (\mathbb Z/p \times \mathbb Z/p^4)$$
makes it straightforward to check that
$$C \cong (\mathbb Z/p^2) \times (\mathbb Z/p^2),$$
with
$$A/C \cong  (\mathbb Z/p^4 \times \mathbb Z/p) \times (\mathbb Z/p^3).$$
Moreover, $pB = pC$. So we have $B \cong C$, $A/B \cong A/C$, and $A/pB \cong A/pC$.
However, there is no automorphism of $A$ which sends $B$ to $C$. For example,
\begin{align}
(pA\cap B)/(p^4A \cap B)
&= (pA_1 \cap \langle b_1 \rangle)/(p^4A_1 \cap \langle b_1 \rangle) \times (pA_2 \cap \langle b_2 \rangle)/(p^4A_2 \cap \langle b_2 \rangle)\\
&\cong 1 \times (\mathbb Z/p^2),
\end{align}
but similarly 
$$(pA\cap C)/(p^4A \cap C) \cong (\mathbb Z/p) \times (\mathbb Z/p).$$
