Consider $f: G\rightarrow\operatorname{Aut}(G)$ where $G=\Bbb{Z}_{10}$. Since this is equivalent to considering $G\rtimes_f G$, we know that $f$ is defined as $f_a(b)=aba^{-1}$.
There are four automorphisms of $G$, which we will say are $\{c_1,c_3,c_7,c_9\}$ defined as $c_k: z\mapsto kz\mod 10$.
Then, $f_a(b)$ is defined as $f_a(c_k)=a+c_k-a$. But since $G$ is abelian, we can then very simply say that $f_a(c_k)=c_k$. Ergo, for every $a\in G, f_a: c_k\mapsto c_k$.
Is this sound reasoning? I keep worrying that I should be taking the multiplicative inverse instead of the additive, but then again, for many elements (like 2), there is no multiplicative inverse.
Additionally, is there a perhaps more instructive example where the automorphism mappings $f_a(b)$ don't just map an automorphism onto itself? Thank you!