Suppose $\phi:\mathbb{Z}_{50} \rightarrow \mathbb{Z}_{15}$ is a group homomorphism with $\phi(7)=3$.
(a) Determine $\phi(x)$
(b) Determine the image of $\phi$
(c) Determine the kernel of $\phi$.
(d) Determine $\phi^{-1}(3)$.
Why is the Kernel such that $5|x$?
On
Hint:
$$\varphi(7) = \varphi(1 + 1 + 1 + \cdots + 1) = \varphi(1) + \varphi(1) + \cdots + \varphi(1) = 7\cdot \varphi(1)$$
Same thing goes for $\varphi(x)$. What is the multiplicative inverse of $7$ modulo $15$?
For your edit: I'm already going to suppose that you've figured out the formula $\varphi(x) \equiv 9x \pmod{15}$. The kernel consists of those $x$ satisfying $\varphi(x) \equiv 0 \pmod{15}$, which amounts to $15$ dividing $9x$. Now, $15$ divides $9x$ if and only if $5$ divides $x$.
If you know $\phi(1)$ you would have it.
In this case it is fairly quick to find $\phi(49),$ where $49$ is the inverse of $1$ in this group