We wish to find all homomorphisms $\phi: \Bbb Z_4 \to \Bbb Z_9^*$.
Now first of all it suffices to know what $\phi(1)$ is since for any other element $m \in \Bbb Z_4$ we have $\phi(m) = \phi(1 + 1 + .. + 1) = \phi(1) + ... + \phi(1) = m \phi(1)$. Hence $\phi(1)$ completely determines $\phi$.
Now, $[\phi(1)]^4 = \phi(1) \cdot \phi(1) \cdot \phi(1) \cdot \phi(1)= \phi(1 + 1 + 1 + 1) = \phi(0) = 1$ since the identity is always mapped to the corresponding identity. Hence, the order of $\phi(1)$ divides $4$. Now we investigate $\Bbb Z_9^*$ for possible candidates to be $\phi(1)$. We notice that only $1$ and $8$ are the elements in $\Bbb Z_9^*$ whose orders are divisible by $4$. So there is at most $2$ homomorphisms, $\phi$.
To establish that there are in fact $2$ homomorphisms you just need to define the map $\phi(x) = 1$ and $\phi(x) = 8^x$ for the two cases.
We wish to find all homomorphisms $\phi: \Bbb Z_4 \to \Bbb Z_9^*$.
Now first of all it suffices to know what $\phi(1)$ is since for any other element $m \in \Bbb Z_4$ we have $\phi(m) = \phi(1 + 1 + .. + 1) = \phi(1) + ... + \phi(1) = m \phi(1)$. Hence $\phi(1)$ completely determines $\phi$.
Now, $[\phi(1)]^4 = \phi(1) \cdot \phi(1) \cdot \phi(1) \cdot \phi(1)= \phi(1 + 1 + 1 + 1) = \phi(0) = 1$ since the identity is always mapped to the corresponding identity. Hence, the order of $\phi(1)$ divides $4$. Now we investigate $\Bbb Z_9^*$ for possible candidates to be $\phi(1)$. We notice that only $1$ and $8$ are the elements in $\Bbb Z_9^*$ whose orders are divisible by $4$. So there is at most $2$ homomorphisms, $\phi$.
To establish that there are in fact $2$ homomorphisms you just need to define the map $\phi(x) = 1$ and $\phi(x) = 8^x$ for the two cases.