Group homomorphism that maps to one element Is there some way to construct a group homomorphism $G \to H$ that maps everything in $G$ to just one non-identity element of $H$ (besides mapping $0$ to $0$)? For arbitrary (finite) $H$ but you have your choice of $G$?
The only example of this is the identity homomorphism $C_2\rightarrow C_2$ (where $C_2$ is the cyclic group of order $2$). Otherwise, if the image of a homomorphism consists of just one non-identity element and the identity, there are multiple elements mapping to the identity.

In particular, let $f:G\rightarrow H$ be a homomorphism. Then, it must hold that, for any $x\in H$ (and letting $1$ be the identity of $H$) such that $f^{-1}(x)$ is not empty:
$$|f^{-1}(x)|=|f^{-1}(1)|$$
- that is, the set of elements of $G$ mapping to $x$ is as large as the set of elements mapping to $1$ under any homomorphism. This is simple to prove, since, if we let $z$ be some element in the preimage $f^{-1}(x^{-1})$ - that is, some element satisfying $f(z)=x^{-1}$ - then it holds that
$$z\cdot f^{-1}(x)=f^{-1}(1)$$
where $z\cdot f^{-1}(x)$ is a coset of $f^{-1}(x)$. This can be proven since, if we apply $f$ to the left side, we would conclude that $f(z\cdot f^{-1}(x))=f(z)x=x^{-1}x=1$ - meaning every element of $z\cdot f^{-1}(x)$ is in $f^{-1}(1)$. Proving the converse - that every element in $f^{-1}(1)$ is in $z\cdot f^{-1}(x)$ is a simple exercise which I will leave to you.

However, what we can prove from this is that if the image of a homomorphism $f:G\rightarrow H$ consists of just two elements, then exactly half of the elements of $G$ map to the identity, and half the elements of $G$ map to the other element. If only one element is permitted to map to the identity, this gives that $1$ is half the order of $G$ - hence $G$ has order two.
Let such a homomorphism $f$ exist and all the non identity elements of $G$ are mapped to the non -identity element $b$ of $H$ .then Im$f$ will consist of $b$ and $e_H$ and ker$f$ will consist only of $e_G$.then by 1st isomorphism theorem

$|G|/|kerf|=|Imf|$

i.e$|G|=|Imf|$ snce ker$f$ consists only of $e_G$

i.e. $|G|=2$ which is the only possibility