Prove that for the semidirect product $\mathbb{Z}_5 \rtimes \mathbb{Z}_3$, the homomorphism $\alpha$ is trivial Suppose $G \cong \mathbb{Z}_5 \rtimes_\alpha \mathbb{Z}_3$ with respect  to a homomorphism $\alpha:\mathbb{Z}_3 \to \mathrm{Aut}(\mathbb{Z}_5)$. Show that $\alpha$ is trivial and that $G \cong \mathbb{Z}_5 \times \mathbb{Z}_3$. Is $G$ a cyclic group?
My proof:
Let $([a]_5, [b]_3), ([c]_5, [d]_3) \in \mathbb{Z}_5 \rtimes_\alpha \mathbb{Z}_3$
The semidirect product gives
$$([a]_5, [b]_3) ([c]_5, [d]_3) = ([a]_5 \alpha_{[b]_3}([c]_5), [b]_3[d]_3)$$
$$= ([a]_5 ([b]_3 [c]_5 [b]_3^{-1}), [b]_3 [d]_3)$$
$$= ([a]_5[c]_5[b]_3[b]_3^{-1}, [b]_3 [d]_3)$$
$$= ([a]_5[c]_5, [b]_3[d]_3)$$
$$= \mathbb{Z}_5 \times \mathbb{Z}_3$$
the definition of the regular direct product.
In the middle of the calculations we see that the inner automorphism is trivial always since $\mathbb{Z}_n$ is abelian. 
Therefore $G \cong \mathbb{Z}_5 \times \mathbb{Z}_3$.
Finally it is cyclic since $\mathrm{gcd}(5,3) = 1$ so $5$ and $3$ are coprime. Therefore the direct product is cyclic
 A: The problem in your reasoning is that you assume $\alpha$ must be
$$\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5): [b]_3 \mapsto \iota_{[b]_3},$$
with
$$\iota_{[b]_3}: \mathbb{Z}_5 \to \mathbb{Z}_5: [c]_5 \mapsto [b]_3[c]_5 [b]_3^{-1}.$$
This "inner automorphism" doesn't make any sense, because you are trying to multiply elements that belong to different groups!
Let us first try to calculate what "known group" $\operatorname{Aut}(\mathbb{Z}_5)$ is isomorphic to, and then try to construct a morphism $\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5)$.
Since $\mathbb{Z}_5$ is cyclic and generated by $[1]_5$, any automorphism $\varphi \in \operatorname{Aut}(\mathbb{Z}_5)$ is determined completely by the value of $\varphi([1]_5)$. There are $5$ possible values of $\varphi([1]_5)$. I'll leave it to you to prove that $\varphi([1]_5) = [0]_5$ will not produce an automorphism, and the other four possibilities will form a group isomorphic to $\mathbb{Z}_4$. Thus $\operatorname{Aut}(\mathbb{Z}_5) \cong \mathbb{Z}_4$.
So we can see $\alpha$ as a morphism 
$$\alpha: \mathbb{Z}_3 \to \mathbb{Z}_4.$$
For the same reason as mentioned above, $\alpha$ is completely determined by the value of $\alpha([1]_3)$. Again, I'll leave it up to you to show the only possible $\alpha$ is
$$\alpha: \mathbb{Z}_3 \to \mathbb{Z}_4: [b]_3 \mapsto [0]_4,$$
which translates to
$$\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5): [b]_3 \mapsto \operatorname{id}.$$
