I found that $\operatorname{Ext}_{\Bbb Z}^1(\Bbb Z_9, \Bbb Z) \cong \Bbb Z_9$ by calculating it as the quotient of two $\operatorname{Hom}$-sets.
I am tasked with finding the (equivalence classes of) exact sequences corresponding to the elements in the group above.
I know two, the trivial one with middle module $\Bbb Z \oplus Z_9$, and the one with middle module $\Bbb Z$ with morphisms $\mu: \Bbb Z \to \Bbb Z$, $\mu(n) = 9n$ and $\phi:\Bbb Z \to \Bbb Z_9$, $\phi(m) = \bar{m}$.
Since $9$ is not prime I don't know what I need to construct the remaining seven exact sequences.
Do you know what a pushout is? The construction you need is
$$ \begin{array}{llllllll} 0 & \to & \mathbb{Z} & \stackrel{\times 9}\to & \mathbb{Z} & \to & \mathbb{Z}/9 & \to & 0 \\ & & \downarrow \phi & & \downarrow & & \downarrow \textrm{id} & & \\ 0 & \to & \mathbb{Z} & \to & P & \to & \mathbb{Z}/9 & \to & 0 \end{array} $$ where $P$ is the pushout of $\times 9$ and $\phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $\mathbb{Z}\oplus \mathbb{Z}$ by the subgroup generated by things of the form $(\phi(z), -9z)$: you can see immediately that $\phi=0$ gives you the trivial extension you mentioned.