Consider a finite abelian group:
$$G \cong \mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$
Let $K$ be a subgroup of $G$, which is not a product of cyclic group, i.e. not of the form: $$K = p_1^{\beta_1}\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times p_2^{\beta_2}\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times p_n^{\beta_n}\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$ with $\beta_i \leqslant \alpha_i.$
I know the example of in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ of $K_1=\langle (0,2)\rangle$ and $K_2=\langle(1,0)\rangle $: $K_1 \cong K_2$ but there is no isomorphism of $G$ that maps $K_1$ to $K_2$. I would be inclined to extrapolate this and think that we cannot prove the existence of an isomorphism of $G$ that maps any subgroup $K$ which is not a product into a product subgroup. If I am right I would need a counter example to prove this non existence assertion. Please answer in this specific case of finite abelian group and for the mapping to "diagonal" to product subgroup.
edit for clarification: Let $$G =\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z} \times\mathbb{Z}/p_2^{\alpha_2}\mathbb{Z} \times\dots\times\mathbb{Z}/p_n^{\alpha_n}\mathbb{Z} $$. Let $K$ be a subgroup of $G$ which is not a product of cyclic groups of prime power orders. Then there exist an isomorphism $\varphi$ of $G$ such that the image of $K$, $\varphi(K)$ is a product of cyclic groups of prime power orders.
I need a counter example to prove the assertion is false. but it might be true; in the latter case i need a full proof or a link to such proof.
I think the subgroup $K=\langle x^2y \rangle$ of order $4$ in the group $G=\langle x,y \mid x^8=y^2=1, xy=yx\rangle$ is a counterexample.
To see that, note that $x^4$ is the only nontrivial fourth power in $G$, so all cyclic subgroups of order $8$ contain $x^4$. But $K$ contains $x^4 = (x^2y)^2$, so $K$ cannot be a direct factor of order $4$. Since $x^2y$ is not a square, it also cannot be contained in a direct factor of order $8$.