Automorphism on integers
Automorphism is a permutation of a set which respects some structure on the set. What structure? It varies. Automorphism is a general term and does not apply simply to groups, or rings.
In the context of $(\mathbb Z,+)$ as an additive group, we say that $f\colon\mathbb {Z\to Z}$ is an automorphism if:
- $f$ is a bijection,
- $f(m)+f(n)=f(m+n)$,
- $f(0)=0$.
Now suppose that $f$ is an automorphism like that. Well, $f(0)=0$. If $f(1)=1$ then $f$ has to be the identity, because for $n>0$ we have $$f(n)=f(\underbrace{1+\ldots+1}_{n \text{ times}})=\underbrace{f(1)+\ldots+(1)}_{n \text{ times}}=\underbrace{1+\ldots+1}_{n \text{ times}}=n$$ and if $n<0$ then $$0=n+(-n)=f(n)+f(-n)=f(n)+(-n)$$ and therefore $f(n)=n$ as well.
Similarly if $f(1)=-1$ then $f(n)=-n$ for all $n$. So these are two automorphisms.
If $f(1)=n$ for some $n>1$, since $f$ is a bijection we have some $k\in\mathbb Z$ such that $f(k)=1$. We can write, if so, $f(1)=f(k)+\ldots+f(k)=f(k+\ldots+k)$. However $f$ is injective and therefore $k+\ldots+k=1$, which can only happen if $k=1$ and $n=1$. In a similar way we show that if $f(1)$ is negative then it has to be $-1$.
All that is well, but we can have more to say here. Suppose now we take $\mathbb Z$ as an ordered group. This means that now we add the ordering $\leq$ and require it plays nice with the addition, as it does with the standard addition and ordering.
In this case it is easy to show that $f(1)=1$ is the only automorphism possible. Why? Because $0<1$ implies that $f(0)<f(1)$, since now $f$ has to preserve the order too, so the first part shows that you cannot have a group automorphism of $\mathbb Z$ except $n\mapsto\pm n$, and if you want order preservation too you can only have the identity.
If on the other hand, you only want to consider $(\mathbb Z,\leq)$ as an ordered set, then you only wish to preserve $n<m\iff f(n)<f(m)$. In this case, it is not hard to show that for every $k\in\mathbb Z$ the function $f(n)=n+k$ is an automorphism.
To sum up, this is really about the structure you wish to preserve. The standard sets you know, $\mathbb {N,Z,R,Q,C}$ are all sets which have several different structures arising naturally from their properties as we know them. Each of these structures has a different property to preserve, and some would allow us rich automorphism groups, while other structures will limit us to have only a few automorphisms, if not one.
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.