In this problem we determine the number of group homomorphisms

$f: \mathbb{Z}_{20} \rightarrow \mathbb{Z}_{50}$

such that the image of $f$ has size exactly $2$.

First, since $20 \equiv 0$ $\pmod{20}$, we have

$$\begin{array}{llllllll} 0 & \equiv & f(0) && \text{property of homomorphisms} \\ & \equiv & f(20) \\ & \equiv & 20 f(1) \pmod{50} && \text{property of homomorphisms.} \end{array}$$
On the other hand, $f(1)$ is an element of $\mathbb{Z}_{50}$ so $0 \equiv 50 f(1) \pmod{50}$.

Thus
$\gcd(20,50) f(1) \equiv 0$ $\pmod{50}$
i.e.
$10 f(1) \equiv 0$ $\pmod{50}$
whence
$(\ast)$ $f(1) = 5 u$ for some $u$ in $\mathbb{Z}_{50}$.

On the other hand, the image of $f$ is a subgroup of the cyclic group $\mathbb{Z}_{50}$. But a cyclic group has at most ONE subgroup of any given order, so if the image of $f$ has size $2$ then the elements of the image of $f$ must be (please enter your answer as an ORDERED list)

Combine this with $(\ast)$ and we see that in order for the image of $f$ to have size $2$, the choices for $f(1)$ are (please enter your answer as an ORDERED list)

Consequently, the number of such functions $f$ is . Please enter your answer as a number.