TITLE: Automorphism on integers
QUESTION [10 upvotes]: Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism? 
If so why does the 2nd example in http://en.wikipedia.org/wiki/Automorphism says that the unique non trivial automorphism is negation?

REPLY [4 votes]: Let $f:\mathbb Z\to \mathbb Z$ be a homomorphism. By definition, for each $m\in \mathbb Z$, $$f(m)=f(1+\cdots+1)=f(1)+\cdots+f(1)=m f(1)=f(1)m$$ Hence any homomorphism $f:\mathbb Z\to \mathbb Z$ is just multiplication by $f(1)$ that is the map $m\mapsto f(1) m$. Now it is easy to see that every such non zero homomorphism is injecitve. Indeed the equation $f(1)m=f(1)n$ implies that $m=n$ since $f(1)\not =0$ by hypothesis and also because $\mathbb Z$ is an integral domain. Now the only cases where these homomorphisms are surjective correspond to when $f(1)=1$ or $f(1)=-1$ hence if we call the case $f(1)=1$ the trivial case then the only remaining case is when $f(1)=-1$ and the corresponding automorphism $m\mapsto -m$ is called the negation.