Consider $H=\Bbb{Z}_{30}$ and $G=\Bbb{Z}_{15}$ as additive abelian groups. Show that ${\rm Aut}(H) \cong {\rm Aut}(G)$.

Consider $$H = \mathbb{Z}_{30}$$ and $$G = \mathbb{Z}_{15}$$ as additive abelian groups. Then how do I show that $${\rm Aut}(H) \cong {\rm Aut}(G)$$?

By the Chinese remainder theorem, I know that $$\mathbb{Z}_{30} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{15}$$. Intuitively I know that the only automorphism of $$\mathbb{Z}_{2}$$ is the identity automorphism, so we can construct a bijective map between $$\psi : {\rm Aut}(G) \rightarrow {\rm Aut}(H)$$. By $$\psi(\phi) = \phi^{*}$$, where $$\phi^{*}((a,b)) = (a, \phi(b))$$, which is an isomorphism. (Here we consider $$\mathbb{Z}_{30} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{15}$$.)

Is the definition of this map sufficient to claim that $${\rm Aut}(G) \cong {\rm Aut}(H)$$? Also, how does one in general find the automorphism group of $$\bigoplus_{k=1}^{r} \mathbb{Z}_{n_{k}}$$, where the $$n_{k}$$'s are not necessarily coprime.

Hint: $${\rm Aut}(\Bbb Z_n)\cong U(n),$$
where $$U(n)$$ is the group of units modulo $$n$$.
For your first question, we have $$\Bbb Z_{30}^×\cong(\Bbb Z_2\times\Bbb Z_{15})^×\cong\Bbb Z_2^×\times\Bbb Z_{15}^×\cong\Bbb Z_{15}^×$$.
And $$\operatorname {Aut}(\Bbb Z_n)\cong\Bbb Z_n^×$$.