What are all the homomorphisms between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$? Any homomorphism $φ$ between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$ is completely defined by $φ(1)$. So from
$$0 = φ(0) = φ(18) = φ(18 \cdot 1) = 18 \cdot φ(1) = 15 \cdot φ(1) + 3 \cdot φ(1) = 3 \cdot φ(1)$$
we get that $φ(1)$ is either $5$ or $10$. But how can I prove or disprove that these two are valid homomorphisms?
 A: Continuing Akhil M's answer above (the comments under it are getting pushed down out of sight): it is also not hard to systematically find all the idempotents in ${\mathbb Z}/n$. Namely, for example with $n=pq$ with distinct primes $p,q$, $\mathbb Z/n \approx \mathbb Z/p \oplus \mathbb Z/q$, by Sun-Ze's theorem (altho' one might carp about what kind of "sum" it is). So, solving the idempotent condition $x^2=x$ mod $pq$ is equivalent to solving that equation mod $p$ and mod $q$. The integers mod a prime form a field, so we know that there are only the two solutions, the obvious ones, $0,1$. Thus, the idempotents mod $pq$ are $0$-or-$1$ mod $p$ and $0-or-1$ mod $q$. Obviously $0$ and $1$ mod $pq$ work, but/and also $0$ mod $p$ but/and $1$ mod $q$, and vice-versa. In the case at hand, both $6$ and $10$ are non-obvious idempotents. 
A: If one has a homomorphism of two rings $R, S$, and $R~$ has an identity, then the identity must be mapped to an idempotent element of $S$, because the equation $x^2=x$ is preserved under homomorphisms.  Now $5$ is not an idempotent element in $\Bbb Z_{15}$, so the map generated by $1 \to 5$ is not a homomorphism.
However, $10$ is an idempotent element of $\Bbb Z_{15}$. In particular, the subring $T \subset \Bbb Z_{15}$ generated by $10$ has unit $10$.  Since it is annihilated by $3$, and consequently by $18$, there is a unital homomorphism $\Bbb Z_{18} \to T$ (i.e., mapping $1$ to $10$).  So your second map is a legitimate homomorphism of rings (composing with the injection $T \to \Bbb Z_{15}$).
Basically, the point of this answer is to check that one of your maps preserves the relations of the two rings, while the other doesn't.
A: I will quote Wikipedia on the definition of a ring homomorphism.

More precisely, if R and S are rings,
  then a ring homomorphism is a function
  f : R → S such that1
  
  
*
  
*f(a + b) = f(a) + f(b) for all a and b in R
  
*f(ab) = f(a) f(b) for all a and b in R
  
*f(1) = 1
  

The last requirement is being relaxed. We can then sub in our f, in this case f(a1)=5a or f(a1)=10a and see if these relations hold. Given that these are small, finite and well understood groups, the problem is easy to solve from here.
