The question asks you to define a binary operation $\ast$ on $\mathbb{Z}$ such that $(\mathbb{Z}, \ast)$ is a group and $\phi : (\mathbb{Z}, +) \to (\mathbb{Z}, \ast)$ is an isomorphism.
Suppose such a binary operation exists. Given $m, n \in \mathbb{Z}$ (thinking of this at the target of $\phi$), how do we define $m\ast n$? Well, by definition of $\phi$, we have
$m\ast n = \phi(m - 1)\ast\phi(n - 1).$
Now, if $\phi$ is an isomorphism from $(\mathbb{Z}, +)$ to $(\mathbb{Z}, \ast)$, we have $\phi(a)\ast\phi(b) = \phi(a + b)$ for all $a, b \in \mathbb{Z}$. In particular, we have
$\phi(m-1)\ast\phi(n-1) = \phi((m-1) + (n-1)) = \phi(m + n - 2).$
Combining the two equations we have
$m\ast n = \phi(m + n - 2),$
but we can simplify this by applying the definition of $\phi$:
$m\ast n = \phi(m + n - 2) = (m + n - 2) + 1 = m + n - 1.$
Therefore, if $\ast$ with the claimed properties exists, then $m\ast n$ must be defined as $m + n - 1$. It is then easy to check that by defining $\ast$ in this way, $(\mathbb{Z}, \ast)$ is indeed a group.