Ring homomorphism in the quaternions I been trying to find a ring homomorphism but without luck, for example in the complex numbers we have the conjugation but this not work in the quaternions, since $f(xy) = f(y)f(x)$ supposing $f$, is the conjugation, and we don't have commutativity in the quaternions.

So, is there some ring homomorphism distinct from the identity or the zero ones? 

 A: If you mean a ring homomorphism from $\mathbb H$ to itself, you can take any noncentral element $x$ and use the ring isomorphism $q\mapsto xqx^{-1}$.
Actually, it can be proven that these (with the identity automorphism) give you all possible automorphisms. 
To show that, we can refer to the Skolem-Noether theorem on central simple algebras. Since $\mathbb H$ is a central simple $\mathbb R$ algebra, we know already that all $\mathbb R$-linear automorphisms are inner.
But if $\phi$ is any automorphism $\mathbb H\to \mathbb H$, then it's easy to see that $r$ is central in $\mathbb H$ iff $\phi(r)$ is central in $\mathbb H$, so that $\phi$ restricted to $\mathbb R$ is an automorphism of $\mathbb R$. But the only automorphism of $\mathbb R$ is the identity, so $\phi$ is already $\mathbb R$-linear. Thus all automorphisms of $\mathbb H$ have been found.
Distinct $x$'s are not guaranteed to yield distinct maps. Since $G=\mathbb H\setminus\{0\}$ is a group, we can refer to the basic group theory result about the conjugation action of $G$ on itself. It's easy to prove that the kernel of the action is the center of $G$ (which corresponds to $\mathbb R\setminus\{0\}$ in our $G$) so that $x$ and $y$ produce the same inner automorphism on $\mathbb H$ iff $x=ry$ for some $r\in\mathbb R$.
