Any example of functions are automorphism? I am looking for functions fulfilling
$f(x+y) = f(x) + f(y)$ and $f(x*y) = f(x)*f(y)$.
I can only find $f(x)=x$, any more?
Any example of functions are automorphism?
 A: The question is not completely stated at this moment, but I thought that it would be better to move my comments to an answer. (Basically they're answering the question in the way I understood it, at least partially. Also, the comments section becomes unreadable after adding too many of them.)
You're asking about functions that simultaneously verify the equations $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$. (You did not specify the domain and codomain, $\mathbb Q\to \mathbb Q$, $\mathbb R\to \mathbb R$ or $\mathbb C\to \mathbb C$ seems to be most plausible. 
You should also include the conditions which you impose on $f$. (I guess at least bijectivity, since you ask about automorphisms in the second part of your question.)

You are correct that $f(x)=x$ is a solution.
Obviously, $f(x)=0$ is a trivial solution for any of these cases, too. (It is not bijective.) If you want to work with functions from $\mathbb C$ to $\mathbb C$, then $f(x)=\overline x$ is a solution.
Here you can find the proof the the only automorphism of the field $\mathbb R$ is trivial, if this was what you intended to ask.
It might be interesting for you to know, that the first equation is called Cauchy functional equation. It is known that the only continuous solutions $f:\mathbb R\to\mathbb R$ are the functions of the form $f(x)=cx$. The proof can be found in some of the questions I linked bellow. 
So if you want such functions to be continuous, the second equation yields $c=1$ or $c=0$. Thus the only continuous solutions $\mathbb R\to \mathbb R$ are $f(x)=x$ and $f(x)=0$.
However, as the above proof shows, if you assume bijectivity and the second equation, the continuity is not needed.

Some related questions (and answers) which you could have a look at:
Field automorphisms of $\mathbb{Q}$ - shouldn't there be only one? 
Are there any non-linear solutions of Cauchy’s equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice? (mathoverflow) 
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t 
