Since $\Bbb{Z}/(p)$ is a ring and $(\Bbb{Z}/p)^{\times} \simeq \Bbb{Z}/(p-1)$ which is also a ring, how do they relate? Let $p$ be an odd prime, then according to: wikipedia article, we have that $(\Bbb{Z}/(p))^{\times} \simeq (\Bbb{Z}/(p-1))^+$.  Then how does multiplication in the ring on the right affect the first ring, and why don't we call the first ring $(\Bbb{Z}/(p), +, \times, *)$, where $*$ is the second multiplication operation?

$\cdot$ on $\Bbb{Z}/(p)$ is a binary map $\mu : \Bbb{Z}/(p) \times \Bbb{Z}/(p) \to \Bbb{Z}/(p)$ written $\mu(x,y) = x\cdot y$.  But multiplication $*$ acts distributively as $z * (x \cdot y) = z * x \cdot z * y = x\cdot x \cdot x (z \text{ times}) \cdot y \cdot y \cdot y (z \text{ times}) = (x\cdot y)^z$ since we're in commutative settings.
 A: The problem with this idea is that $*$ depends on the precise isomorphism used between the $(\mathbb{Z}/(p))^\times$ and $(\mathbb{Z}/(p-1))^+$.  For example, in the latter ring, $0$ and $1$ both square to themselves.  The preimage of $0$ is $1$, but the preimage of $1$ is any (non-1) element of $(\mathbb{Z}/(p))^\times$.

More explanation, per request.  One way to establish the isomorphism $\phi$ between $\mathbb{Z}_7^\times$ and $\mathbb{Z}_6^+$ is to send $2$ to $1$.  Now, in the latter ring, $1\times 1=1$; hence, with this new operation, $2*2=2$.  In the first ring, $4=2\times 2$, hence $\phi(4)=\phi(2)+\phi(2)=1+1=2$.  Hence, $4*4=\phi^{-1}(2\times 2)=\phi^{-1}(4)=\phi^{-1}(1+1+1+1)=2\times 2\times 2\times 2=16=2$.

But, a different way to establish the isomorphism is to send $4$ to $1$.  Now, again $1\times 1=1$ in the latter ring, so $4*4=4$.

In short, what you need is actually $(\mathbb{Z}/(p),+,\times,\phi,*)$, where $*$ depends on $\phi$.  This makes $*$ not particularly natural.  If you were to find interesting properties of $*$, people might care, but otherwise it's just another potential binary operation.
