I am trying to find $\text{Aut}(\mathbb Z_2^3)$ and express it in terms of familiar groups and the direct and/or semi direct product.
Here's what I have so far: I know that the set of generators $A:=\{(1,0,0),(0,1,0),(0,0,1)\}$ of $\mathbb Z_2^3$ must be mapped onto another set of $3$ generators by an arbitrary automorphism $\varphi$. I have calculated combinatorially that $\mathbb Z_2^n$ has $2^n-1$ sets of $n$ generators, and since each automorphism maps $A$ onto one of these $n$-element sets in one of $n!$ ways, there are a total of $n!(2^n-1)$ automorphisms of $\mathbb Z_2^n$.
In summary, $|\text{Aut}(\mathbb Z_2^n)|=n!(2^n-1)$. So for $n=3$, we can see that $|\text{Aut}(\mathbb Z_2^3)|=6\cdot 7=42$.
Is this reasoning all correct? If so, how do I proceed from here to actually find out what the group $\text{Aut}(\mathbb Z_2^3)$ is?