I want to list all the subgroups of the semi-direct product $\mathbb{Z}/7\mathbb{Z} \rtimes (\mathbb{Z}/7\mathbb{Z})^{\times}$, under the homomorphism $\theta: (\mathbb{Z}/7\mathbb{Z})^{\times} \rightarrow \mathrm{Aut}(\mathbb{Z}/7\mathbb{Z})$, $\theta: a \mapsto \theta_{a}$ where $\theta_{a}(i)=ai$. Until now, I know that the subgroups of $(\mathbb{Z}/7\mathbb{Z})^{\times}$ will be of orders $1, 2, 3$ or $6$ and moreover they will be unique (similarly, the cyclic group with $7$ elements only has the trivial subgroups).
I was thinking that the subgroups of the semi-direct product would be semi-direct products of the subgroups of $\mathbb{Z}/7\mathbb{Z}$ and $(\mathbb{Z}/7\mathbb{Z})^{\times}$. Is my claim correct? If not, what would be a way to compute those subgroups?
Let $H$ be a subgroup of the group $G$ of order $42$ in question.
If $7$ divides $|H|$, then, since $G$ contains normal subgroup $P$ of order $7$ which is Sylow-$7$ subgroup, $H$ will contain the normal subgroup $P$. Then $H/P$ is subgroup of $G/P\cong \mathbb{Z}_7^{\times}$; so possible orders of $H/P$ are $1,2,3,6$ and it is unique (in $\mathbb{Z}_7^{\times}$), according to which we will get unique subgroups of $G$ of order $7,14,21,42$.
If $7$ does not divide $|H|$ then $|H|$ will be $1,2,3$ or $6$, $H$ will be conjugate to subgroup of $\mathbb{Z}_7^{\times}$; this is because of the following: if $|H|=1$ then it is obvious. If $|H|=6$, then $H$ is complement of a normal Hall subgroup $\mathbb{Z}_7$ and by Schur-Zassenhaus, the complements of a normal Hall subgroup are conjugate. If $|H|=2$ or $3$ then $H$ will be Sylow-$2$ or Sylow-$3$ subgroup, and $\mathbb{Z}_7^{\times}$ contains Sylow-$2$ and Sylow-$3$ subgroup, hence $H$ will be conjugate to a subgroup of $\mathbb{Z}_7^{\times}$.