Is $( \mathbb{ Z}^*,\cdot) \rightarrow (\mathbb{Z}_5^{*},\cdot), n \mapsto n \pmod 5 $ well-defined?
So what I think is that it is not well -defined because the non-zero multiples of 5 in $\mathbb{ Z}^*$ map to $[0]_5$ but $[0]_5 \notin \mathbb{Z_5^*}$
What do you think?
You seem to think that $\mathbb{Z}^*=\mathbb{Z}\backslash\{0\}$. That would be a rather uncommon notation and it is likely to be wrong. At least in pure algebra.
Given a unital ring $R$, typically $R^*$ denotes the group of all invertible elements of $R$:
$$R^*=\big\{x\in R\ \big|\ xy=yx=1\text{ for some }y\in R\big\}$$
which is a group under multiplication inherited from $R$. Note that $0\not\in R^*$ unless $R$ is a trivial (one element) ring. And so $R^*\subseteq R\backslash\{0\}$, but $R^*$ may and often is a proper subset of $R\backslash\{0\}$. In fact $R^*=R\backslash\{0\}$ if and only if $R$ is a division ring (or field in commutative case). So for example $\mathbb{Z}_5^*$ is equal to $\mathbb{Z}_5\backslash\{0\}$, but only because $\mathbb{Z}_5$ is a field.
But $\mathbb{Z}$ is not a division ring and in fact $\mathbb{Z}^*=\{-1,1\}$. It is literally the set containing two elements: $-1$ and $1$, no other integer is invertible in $\mathbb{Z}$. So multiplicies of $5$ are not even in $\mathbb{Z}^*$. You can easily verify that with this $n\mapsto n\text{ mod }5$ is a well defined group homomorphism.
In fact if $f:R\to S$ is any unital ring homomorphism, then it induces a group homomorphism $f^*:R^*\to S^*$ given by $f^*(x)=f(x)$. We can apply this to the well known unital ring homomorphism $\mathbb{Z}\to\mathbb{Z}_m$, $n\mapsto n\text{ mod }m$.
Edit: Unless you actually do define $\mathbb{Z}^*$ as $\mathbb{Z}\backslash\{0\}$. In that case your reasoning is correct and this is not a well defined function.