Characterization of the isomorphic semidirect products Let $A$ and $G$ be two finite abelian groups and let $\alpha$, $\beta:G\rightarrow{\rm Aut}(A)$. Suppose that $\alpha (G)$ and $\beta (G)$ are conjugate subgroups of ${\rm Aut}(A)$. Are the semidirect products $A\rtimes _{\alpha }G$ and $A\rtimes _{\beta }G$ isomorphic?
I know that this is true for a finite cyclic group $G$ but I don't what to do if $G$ is a finite non-cyclic Abelian group. I think the answer is usually no, so I will be thankful if someone provides me a counterexample.
Thank you in advance.

That is not the answer to the question. Here is a condition under which the named groups are isomorphic.
If $f\in\operatorname{Aut} A$ is chosen such that for any $x\in G$ and $a\in A$ holds
$$
f^{-1}\alpha(x)f(a)=\beta(x)(a)
$$
and multiplications in the group $A\rtimes_\alpha G$ are given by the rule
$$
(a,x)(b,y)=(a+\alpha(x)(b),xy)
$$
(we use the additive notation of the operation in $A$),
then the mapping $(a,x)\to (f(a),y)$ is an isomorphism of the group $A\rtimes_\alpha G$ to the group $A\rtimes_\beta G$.
