1) Determine the Cayley Table of $(Z_5^*, \cdot)$
2) determine which additive group has the exact same table.
3) Further determine an isomorphism between those two groups and prove by means of that isomorphism that $(Z_5^*, \cdot)$ is a group
So since 5 is a prime number, each element different than 0 has a muliplicative inverse element and the Cayley table may be written as (is this correct?) $$\begin{array}{c|c|c|c|c} 5^{*} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ \hline 2 & 2 & 4 & 1 & 3 \\ \hline 3 & 3 & 1 & 4 & 2 \\ \hline 4 & 4 & 3 & 2 & 1 \end{array}$$
So we see that [4] is its own multiplicative inverse and [2] and [3] are the multiplicative inverses of each other. However I have no idea what to do about questions 2) and 3).
$\mathbb Z/4\mathbb Z, +$ is isomorphic to $(\mathbb Z/5\mathbb Z)^\times,\times$.
Both have two elements of order $4$: $1,3\in\mathbb Z/4\mathbb Z$ and $2,3\in(\mathbb Z/5\mathbb Z)^\times$.
An isomorphism will map one of these elements in $\mathbb Z/4\mathbb Z$ to one of those elements in $(\mathbb Z/5\mathbb Z)^\times$ .
So there are actually two isomorphisms from $\mathbb Z/4\mathbb Z$ to $(\mathbb Z/5\mathbb Z)^\times$ :
one maps $1\mapsto2$, and the other maps $1\mapsto3$.
Since $1$ generates the elements of $\mathbb Z/4\mathbb Z$, once the image of $1$ is determined,
the images of all the elements are determined; e.g., $\phi(3)=\phi(1+1+1)=\phi(1)^3$.
Determining the isomorphism in this way ensures that $(\mathbb Z/5\mathbb Z)^\times,\times$ is a group, just like $\mathbb Z/4\mathbb Z, +$.
To write the isomorphisms the other way around, $\Psi(2)=1$ or $3$ and $\Psi(2^n)=n\Psi(2)$.