What is the relation between extending $\varphi_a$ to an isomorphism and finding the generators of $\mathbb Z_{48}$?

Question

here is the question I wanna understand its solution:

Let $\mathbb Z_{48} = \langle x \rangle.$ For which integers does the map $\varphi_a$ defined by $$\varphi_a: \bar{1} \rightarrow x^a,$$ extends to an isomorphism from $\mathbb Z/48 \mathbb Z$ onto $\mathbb Z_{48}.$

Here is a solution I found online:

From proposition 6(2) on pg. 57 in Dummit and Foote (third edition) , we know that if $H = \langle x \rangle$ and $|x| = n < \infty.$ Then $\langle x \rangle = \langle x^a \rangle$ iff $\operatorname{gcd}(a,n)=1.$ In particular, in our case here, let $H = \mathbb Z_{48} = \langle x \rangle,$ so $x^a$ is a generator of $\mathbb Z_{48}$ precisely when $\operatorname{gcd}(a,48)=1.$\

But, for the sake of simplicity of calculations, we know that $48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3.$ So the generators of $\mathbb Z_{48}$ are $x$ raised to the power of those integers which are not multiples of $2$ and $3.$ Thus the generators of $\mathbb Z_{48}$ are $x^1, x^5, x^7, x^{11}, x^{13}, x^{17}, x^{19}, x^{23}, x^{25}, x^{29}, x^{31}, x^{35}, x^{37}, x^{41}, x^{43}, x^{47}$Thus $\varphi_a$ is an isomorphism for those values of $a.$

But the following questions triggers my mind:

1- What is the importance of the domain of $\varphi_a$ being 1?

2- Are we seeking finding the generators of $\mathbb Z_{48}$ or $\mathbb Z/48 \mathbb Z$ and why?

3- why finding the generators will extend $\varphi_a$ to an isomorphism? what is the relation between them?

4- Is the above solution a complete solution to the problem?

Could anyone help me answer those questions please?

Thanks in advance!

EDIT:

An answer to at least one question of my questions is fine and I will post the rest of my questions in another post.

EDIT:

I found also the solution below online but I doubt it is correct as it did not give me the required values of a:

Any clarification to the correct steps of the solution will be greatly appreciated.

Answer 1

1. The element 1 generates the group, therefore, knowing $\phi_a$ of that element uniquely determines a homomorphism of the whole group. One could have chosen other generators of the group (all numbers coprime to 48).

2. Here, the generators are found for $\mathbb{Z}_{48}$.

3. See what I answered to (1). If you have a generator $x^a$ for the group, then every element of $\mathbb{Z}_{48}$ can be written as a power of this element. Therefore, the elements $x^a, (x^a)^2, (x^a)^3, ..., (x^a)^{48}$ are all distinct. But since $(x^a)^b= (x^b)^a$, those elements can be written as $\phi_a(1), \phi_a(2), ,..., \phi_a(48)$, so $\phi_a$ is surjective, and therefore (since $\mathbb{Z}_{48}$ is finite) injective, and hence an isomorphism.

4. Yes, the solution is fine.

Your questions tells me that you find the whole thing very confusing, so as someone said in the comments, you should probably make a thread for each of them, so the answers will maybe be more thorough than mine here.