So im trying to map $1$ (a generator of $\Bbb{Z}_{45}$) to an element in $\Bbb{Z}_{36}$. So what i did here was that, i know $|f(a)|$ must divide $|a|=45$, and also $|f(a)|$ must divide $|\Bbb{Z}_{36}|=36$; this narrows the possibilities of $|f(1)|$ to be $1, 3, 9$. Now since i have these, i know that the elements in $\Bbb{Z}_{36}$ that satisfy this will be such that either they are $36/1$ or $36/3$ or $36/9$, or $k$ such that $\gcd(k,36)=1,3,9$.
Now, correct me if im wrong in any of my assumptions above. The way I found my answers felt strange; do I have to just think of all possible numbers in $\Bbb{Z}_{36}$ that might satisfy these conditions? It feels like theres a good amount that do, which gives me the feeling I'm doing something wrong here.
From the Isomorphism Theorem you got correctly, that the cardinality of image must be a common divisor of $36$ and $45$, hence only $9, 3$ and $1$.
If the image is of cardinality $1$, it means that $f(1) = 0$, done.
Now remember/realize that for any divisor $k$ of $n$, there is precisely one subgroup of $\mathbb{Z}_n$ of cardinality $k$ and this is isomorphic to $\mathbb{Z}_k$, hence it has $\varphi(k)$ distinct generators. Let's use this:
if the image is of cardinality $3$, then it must be isomorphic to $\mathbb{Z}_3$ (actually the subgroup with elements {0, 12, 24}), which has $\varphi(3) = 2$ generators (elements 12 and 24), so $f$ can map $1$ to any of them, hence 2 homomorphisms.
Similarly for the image of cardinality $9$ you get $\varphi(9) = 6$ different homomorphisms.
In total it looks like 9 homomorhisms.