Show that $\mathbb{Z}$ is isomorphic to $n\mathbb{Z}$. Could someone please see whether my solution is okay? I cannot find any duplicates, since my search engine on here doesn't let me search well.

Show that $\mathbb{Z}$ is isomorphic to $n\mathbb{Z}$ for $n\neq 0$.

Define $\phi:\Bbb{Z}\to n\Bbb{Z}$ by $\phi(r)=nr$ for some fixed $n\in \Bbb{Z}^{*}$.
Well-defined: If $r=s$ for $r,s\in \Bbb{Z}$, then $\phi(r)=nr=ns=\phi(s)$.
Operationing-preserving: $\phi(r+s)=n(r+s)=nr+ns$ (integers distribute over addition) $=\phi(r)+\phi(s)$
Injective: If $\phi(r)=\phi(s)$, then $nr=ns$ or $r=s$ since $n\neq 0$.
Surjective: $\phi$ is onto iff for all $nr\in n\Bbb{Z}$, there exists an $r\in \Bbb{Z}$ such that $\phi(r)=nr$. By construction of $\phi$, it is onto.
 A: This is correct, with one odd comment: your "well-definedness" step is unnecessary.

Since this can be a subtle point, let me say a bit about what's going on here. "Well-definedness" shows up in the following situation:

Suppose we have a set $E$, and elements of $E$ correspond to equivalence classes under $\sim$ of some set $S$. Then we can define a function from $E$ by defining a function from $S$ instead ... provided that we separately show that $\sim$-equivalent elements of $S$ get sent to the same object.

I've deliberately written that informally. More precisely, but also maybe less intuitively:

Suppose $E=S/\sim$. Then if $f:S\rightarrow X$ has the property that $s_1\sim s_2\implies f(s_1)=f(s_2)$, then $f$ "induces" a function $\hat{f}: E\rightarrow X$ given by $\hat{f}([s])=f(s)$ for all $s\in S$.

Let's consider an example. We can think of a rational number as an equivalence class of elements of $\mathbb{Z}\times \mathbb{Z}_{\not=0}$, given by $$(a, b)\sim (c, d)\iff ad=bc.$$ For example, $(1, 2)\sim (2, 4)$ - that is, ${1\over 2}={2\over 4}$. Now consider the following:

Let $f:\mathbb{Q}\rightarrow\mathbb{Z}$ be defined by $$f({a\over b})=a+b.$$ 

(Here I suggestively write "${x\over y}$" instead of "$(x, y)$" for simplicity.) This is of course nonsense. What's really going on is:


*

*I've defined a function $f:\mathbb{Z}\times\mathbb{Z}_{\not=0}\rightarrow\mathbb{Z}$ given by $f(a, b)=a+b$.

*I've then implicitly claimed, without justification, that $f$ "respects $\sim$" - that is, $(a, b)\sim (c, d)\implies f(a, b)=f(c, d)$. 
This second step of course is false: as observed above, $(1, 2)\sim (2, 4)$, but I leave it as an exercise to show that $1+2\not=2+4$.
Now what does this have to do with your situation? Well, nothing! That's because you've defined a function from $\mathbb{Z}$ directly; you're not defining a function on a set which $\mathbb{Z}$ is a quotient of, and then implicitly claiming that that function respects the equivalence relation. So there's no "well-definedness" claim here that needs to be proved.
(OK, there is actually a claim that needs to be proved here: you need to prove that the map $r\mapsto nr$ is actually a function from $\mathbb{Z}$ to $n\mathbb{Z}$, that is, each $x\in\mathbb{Z}$ gets sent to some $y\in n\mathbb{Z}$. This is essentially trivial - obviously $nr\in n\mathbb{Z}$, as long as $r\in\mathbb{Z}$ - but is necessary to prove in the sense of providing a completely rigorous proof. But that's not what you're talking about when you say "well-definedness," and also might not be something you need to explicitly write in this particular exercise.)
A: As Serge Lang wrote at the beginning of his Algebra book:

Let $A$ be a set with an equivalence relation, let $E$ be an equivalence class of elements of $A$. We sometimes try to define a map of the equivalence classes into some set $B$. To define such a map $f$ on the class $E$, we sometimes first give its value on an element $x \in E$ (called a representative of $E$), and then show that it is independent of the choice of representative $x \in E$. In that case we say that $f$ is well defined.

This is the context in which "well defined" comes up. If your equivalence relation is just "$=$" then checking if something is well defined is the same as checking if it is defined. After you define something, hopefully, you already know whether or not that thing is defined.
Also $\mathbb{Z}^* = \{-1,1\}$ is the set of invertible integers. Only in a field $F$ do you have $F^* = F \setminus \{0\}$.
The rest is good.
