If $f$ is an automorphism of $Z$, and $I$ is the identity function on $Z$, show that f is either $I$ or $-I$ If $f$ is an automorphism of $Z$, and $I$ is the identity function on $Z$, show that f is
either $I$ or $-I$

Def of Automorphism is that it maps to itself $f:Z \to Z$
def of function 


*

*$f:X \to Y$ $\forall x\in X, \exists y \: f(x)=y$

*well defined



$1_i:Z \to Z$ maps to $Z$ where $1_i(z)=z*1$ again for evry $Z$ and is well defined. It is true for $-1_i: Z \to Z$ where $-1_i(z)=z*(-1)$
Is it proved by contradiction???
If the mapping is not $I $ or $-I$ then there is a contradiction??
 A: I'm going to assume you mean group automorphisms of $\mathbb{Z}$, i.e. bijections $f : \mathbb{Z} \to \mathbb{Z}$ which are also group homomorphisms.
The easiest way to prove this is first to look at the family of group endomorphisms on $\mathbb{Z}$, i.e. group homomorphisms $\mathbb{Z} \to \mathbb{Z}$. I claim that if $f : \mathbb{Z} \to \mathbb{Z}$ is a group homomorphism, then $f(x) = f(1) \cdot x$ for all $x \in \mathbb{Z}$. The idea here is that if I know $f(1)$, then I know, say, 
\begin{align*}
f(5) & = f(1 + 1 + 1 + 1 + 1) \\
& = f(1) + f(1) + f(1) + f(1) + f(1) \\
& = 5 \cdot f(1) .
\end{align*}
Basically, I can "build" $\mathbb{Z}$ from $1$; consequently, I can build any homomorphism out of $\mathbb{Z}$ from $1$ as well.
From here, once we know that the group homomorphisms from $\mathbb{Z}$ to itself are all just equivalent to multiplying by some integer, call it $C$, we can ask when a morphism is bijective. Here, we need only observe that if $f(x) = C x$, then $f(x) = y$ if and only if $y / C = x$. But this means that $1$ is only in the image of $f$ if $1/C \in \mathbb{Z}$, which occurs only if $C = \pm 1$. Thus if $x \mapsto Cx$ is onto, then $C = \pm 1$.
As others have said, in the future, pay attention to what kind of homomorphism you're dealing with. Context will often make it clear, but a little clarity never hurts, especially when you first get into a topic.
A: The kicker, here, is that anything that says "morphism" should probably have a context. So far, you haven't provided any, so it could easily be a set automorphism--i.e.: a bijective function--such as $f(n)=n+1$.
However, suppose we require more. For example, suppose we require that $f$ be a group automorphism--i.e.: a bijective function satisfying the following:


*

*$\forall x,y\in\Bbb Z,$ $f(x+y)=f(x)+f(y),$

*$f$ is a bijection $\Bbb Z\to\Bbb Z.$


Then one can show the desired results.
A: I take it that by "automorphism" our OP Iloveass2 means "ring automorphism", in which case we have:
The only non-trivial ring homomorphism $f:\Bbb Z\to \Bbb Z$ is the identity map $I$.
For let $f:\Bbb Z \to \Bbb Z$ be any ring homomorphism.  Then
$f(1) = f(1^2) = f(1)f(1), \tag 1$
which may only be satisfied if
$f(1) = 0 \tag 2$
or
$f(1) = 1; \tag 3$
in the former case for any $n \in \Bbb Z$ we have
$f(n) = f(n \cdot 1) = f(n) \cdot f(1) = f(n) \cdot 0 = 0, \tag 4$
so $f$ is in fact trivial contrary to our hypothesis, and thus we rule out case (2); if $f(1) = 1$, then
$f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2, \tag 5$
$f(3) = f(2 + 1) = f(2) + f(1) = 2 + 1 = 3; \tag 6$
if $0 < k \in \Bbb Z$ and
$f(k) = k, \tag 7$
then
$f(k + 1) = f(k) + f(1) = k + 1; \tag 7$
thus we see that 
$f(n) = n \tag 8$
for all $n \in \Bbb N$.  Also,
$f(0) = f(0 + 0) = f(0) + f(0), \tag 9$
whence
$f(0) = 0, \tag{10}$
and since for $n \ge 0$
$n + (-n) = 0, \tag{11}$
$f(n) + f(-n) = f(n + (-n)) = f(0) = 0, \tag{12}$
so
$f(-n) = -f(n) = -n, \tag{13}$
and we see $f$ fixes the negative integers as well as the non-zero ones. It follows that
$\forall n \in \Bbb Z \; f(n) = n, \tag{14}$
so we conclude 
$f = I, \tag{15}$
the identity map on $\Bbb Z$.
I think it is worth noting that the only place we invoke multiplication in the above is to show $f(1) = 1$.  Once we have that, the rest is all about addition.  One might even say,
"God made One; all else, the work of man!"
