Question: A homomorphism from $G$ to itself is an automorphism if it is bijective. 

I am trying to make the condition of bijectiveness weaker. 1-1 is not enough because there is a 1-1 homomorphism from $\mathbb{Z}$ to $\mathbb{2Z}$. What about onto? If a homomorphism from $G$ to itself is onto, then is it an automorphism? Or, similarly, if $H$ is a nontrivial normal subgroup of $G$, can $G$ and $G/H$ be isomorphic?

Answer: 
Consider $z\mapsto z^n$ from the group of nonzero complex numbers to itself.
By fundamental theorem of algebra it is onto. That is, any complex number has $n$th roots. But it takes the same value on all $n$th roots of unity. So an infinite group quotiented by a finite subgroup CAN BE isomorphic to itself.
If $G$ is finite, yes, for one-one and onto are equivalent.

For you last question, $S^1/C\_n\simeq S^1$.