If $G$ is infinite cyclic group, then identity map and inverse map is the only isomorphism. Let $\varphi:G\to G$ . and $\varphi(g) = g^{-1}$ which I said as inverse map.
First, I proof $\varphi$ is an isomorphism by checking bijection and homomorphism.
and I know the theorem which said the inverse of isomorphic map is also isomorphic. Thus I said identity map is isomorphic.
I noted that $G$ is infinite cyclic group, so it is isomorphic to $\Bbb Z$.
So I thought the operation on $G$ should be addition "+".
Then I conclude the theorem that there is no other function that can goes from $G$ to $G$.
But I need some exact proof and want to know it's wrong or not.
 A: Because the group is cyclic, $G=\langle g\rangle$, we have $f(G) = \langle f(g)\rangle$. Since it is a homomorphism, if $f(g) = kg$ then

$$f(ng) = nf(g) =nkg.$$

In order for this to be surjective, $g\in f(G)$ hence $nk=1$ for some $n$ and therefore $k|1$ by definition, but then $k=\pm 1$ is necessarily the case.
A: Let's start with a group $G=\langle g\rangle$. Assuming that you prove that the inverse map (in this case, using addition it would be $f(g)=-g$), as you've stated, what remains is to prove that there are no other such maps. 
A typical way to do this is to assume that you have an isomorphism and then prove that it must be one of the two functions that you've already identified.
Let $h:G\to G$ be an isomorphism. Then $h(g)=ng$ for some $n\in \mathbb{Z}$ (because $G$ is generated by $g$). Furthermore, $h^{-1}(g)=mg$ for some $m\in\mathbb{Z}$ (for the same reason). Then \begin{align} g&=h^{-1}(h(g))\\&=h^{-1}(ng)\\&=nh^{-1}(g)\\&=nmg \end{align}
Since $G$ is infinite, $nm=1$, which immediately implies that $n=m=1$ or $n=m=-1$ and $h$ is one of the previously identified isomorphisms.
A: Let $u$ be a generator of $G$. Then the only other generator is $-u$.
If $\phi$ is an automorphism of $G$, then $\phi(u)$ must be a generator and so $\phi(u)=\pm u$.
Therefore, $\phi$ is the identity map or the inverse map.
