Q:

Prove that $G$ is either the trivial group $\{1\}$ or isomorphic to $\mathbb Z / 2 \mathbb Z$.

Problem: Suppose $G$ is a finite abelian group and there is no non-trivial isomorphism $f: G \rightarrow G$ which has order $2$. Prove that $G$ is either the trivial group $\{1\}$ or isomorphic to $\mathbb Z / 2 \mathbb Z$.
My idea:
Let $f: G \rightarrow G, x \mapsto x^{-1}$, we have
$$f(xy)=(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}=f(x)f(y), \forall x,y \in G.$$
Also, $\ker f=\{x \in G: x^{-1}=1\}=1.$ For all $x\in G$, there exists $x^{-1} \in G$ s.t $(x^{-1})^{-1}=x$. Thus, $f$ is isomorphism. 
$f$ has order $2$ since $f(f(x))=x, \forall x \in G$, in other words, $f\circ f=\rm id$. 
Therefore, $f$ must be trivial, $f=\rm id$, it induces $x^2=1, \forall x\in G$.
Assume that $X\ne \{1\}$, I want to prove that $|X|=2$, but I have no idea. 
Help me, thanks! 

A:

I'll write $G$ additively, rather than multiplicatively.  As you say, the automorphism $x\mapsto -x$ has order $2$ unless $G$ is $2$-torsion, that is
$x+x=0$ for all $x\in G$.
A $2$-torsion Abelian group is effectively a vector space over the field $\Bbb F_2$, so has the form
$$G=\Bbb F_2^k=\{(a_1,\ldots,a_k):a_1,\ldots,a_k\in\Bbb F_2\}.$$
When $k\ge2$, $G$ has the non-trivial automorphism
$$\phi:(a_1,\ldots,a_k)\mapsto(a_1+a_2,a_2,\ldots,a_k)$$
which is its own inverse.  So unless $k\le 1$, $G$ has an automorphism of order $2$.

