I know that $\mathbb{Z}_2=\{0,1\}$ and $\mathbb{Z}_4^*=\{1,3\}$
So $\phi$ is defined as:$\phi(0)=1$ and $\phi(1)=3$
Clearly then it is bijective since every element of $\mathbb{Z}_2$ maps to exactly one unique element in $\mathbb{Z}_4^*$ and vice versa.
However to show that it is an isomorphism it needs to satisfy: $\phi(a*b)=\phi(a)*\phi(b)$, which in this case it does not since $\phi(0*1)=\phi(0)=1\neq\phi(0)*\phi(1)$. Am I missing something or misunderstand something? Any nudge in the correct direction is greatly appreciated.
As the comments have noted, $\mathbb{Z}/(2)$ is an additive group and $(\mathbb{Z}/(4))^*$ is a multiplicative group. So, you want $\phi(a+b) = \phi(a)\phi(b)$. You have $$\phi(0+1) = \phi(1) = 3 = \phi(0)\phi(1),$$ $$\phi(0+0) = \phi(0) = 1 = \phi(0)\phi(0),$$ $$\phi(1+1) = \phi(0) = 1 = \phi(1)\phi(1).$$