Suppose that we have a group $A$ containing a normal subgroup $G$ and two complements $H$ and $K$.
Symbolically, $H, K \leq A$; $G \unlhd A$; $GH = GK =A$; $G \cap H = G \cap K = \{1\}$; so that $A \cong G \rtimes H$ and $A \cong G \rtimes K$.
Can I say that there is an automorphism of $A$ sending $H$ to $K$? Could this automorphism also fix $G$? What if $G$ is furthermore supposed characteristic in $A$? I can't see a simple way to prove it.
Thanks in advance.
The answer is no. Let $A = G \times H$ with $G = S_3$ and $H =C_2$. So $H \le Z(A)$, and we can take $K$ to be a noncentral complement of $G$ in $A$.
The answer is still no if $G$ is characteristic in $A$. Take $G=A_5$ and $H=C_2$.
The answer is yes when $G$ is abelian, because then $H$ and $K$ must induce the same action $\phi$ on $G$.