Automorphism group of a semi-direct product

I'm trying to construct the semi-direct product $(\mathbb{Z}_7 \rtimes \mathbb{Z}_3) \rtimes \mathbb{Z}_2$. Constructing the first factor in parentheses is not difficult. But when it comes to constructing the entire semi-direct product I'm having issues finding the maps $\phi: \mathbb{Z}_2 \rightarrow \text{Aut}(\mathbb{Z}_7 \rtimes \mathbb{Z}_3)$. First, with regard to the semi-direct product which I mentioned in the beginning, does such a thing actually exist, or should it be $(\mathbb{Z}_7 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2$ or $(\mathbb{Z}_7 \rtimes \mathbb{Z}_3) \times \mathbb{Z}_2$? Also, I'm not so familiar with automorphism groups of semi-direct products, which I believe is the reason I'm not getting very far. I would appreciate some help here.

• Are you sure about that? I thought the automorphism group of a cyclic group of order $p$ for $p$ prime has order $\phi(p)= p-1$. – Libertron Jan 3 '15 at 18:25
• Yes, sorry, I was careless. But then the hint should have been to look at the semidirect product of a cyclic group of order $7$ with its automorphism group of order $6$. – Geoff Robinson Jan 3 '15 at 18:39
• Well, I did use that hint to construct $\mathbb{Z}_7 \rtimes \mathbb{Z}_3$, but what is the order of $\text{Aut}(\mathbb{Z}_7 \rtimes \mathbb{Z}_3)$? – Libertron Jan 3 '15 at 18:43
• Well, the point I was trying to make is that if you look at the semidirect product a cyclic group of order $7$ with a cyclic group of order $6$ (inducing an automorphism genuinely of order $6$), there is a normal subgroup of order $21$. – Geoff Robinson Jan 3 '15 at 18:46
• [spoiler] One such group would be $\mathsf{AGL}(1,7)$, i.e. the maps $\Bbb F_7 \mapsto \Bbb F_7 : x \mapsto ax+b$, where $b\in\Bbb F_7$ and $a\in\Bbb F_7^*$. (This is isomorphic to the group that Geoff Robinson is hinting toward.) – Myself Jan 3 '15 at 19:54