Let us first consider the case when $G$ is a non-abelian group.
Then there exist elements $g, h\in G$ such that $gh\neq hg$.

Consider the map $\phi: G \to G$ defined by sending $x\in G$ to $gxg^{-1}$.
Then it is straightforward to check that $\phi$ is a group homomorphism and its inverse is given by the conjugation by $g^{-1}$.
Hence $\phi$ is an automorphism.

If $\phi=1$, then we have $h=\phi(h)=ghg^{-1}$, and this implies that $gh=hg$.
This contradicts our choice of $g$ and $h$.
Hence $\phi$ is a non-trivial automorphism of $G$.

Case When $G$ is an Abelian Group

Next consider the case when $G$ is a finite abelian group of order greater than $2$.
Since $G$ is an abelian group the map $\psi:G\to G$ given by $x \mapsto x^{-1}$ is an isomorphism, hence an automorphism.

If $\psi$ is a trivial automorphism, then we have $x=\psi(x)=x^{-1}$.
Thus, $x^2=e$, where $e$ is the identity element of $G$.

Sub-Case When $G$ has an Element of Order $\geq 3$.

Therefore, if $G$ has at least one element of order greater than $2$, then $\psi$ is a non-trivial automorphism.

Sub-Case When Elements of $G$ has order $\leq 2$.

It remains to consider the case when $G$ is a finite abelian group such that $x^2=e$ for all elements $x\in G$.
In this case, the group $G$ is isomorphic to
\[\Zmod{2}\times \Zmod{2}\times \cdots \Zmod{2}=(\Zmod{2})^n.\]
Since $|G| > 2$, we have $n>1$.

Then the map $(\Zmod{2})^n\to (\Zmod{2})^n$ defined by exchanging the first two entries
\[(x_1, x_2, x_3, \dots, x_n) \mapsto (x_2, x_1, x_3, \dots, x_n)\]
is an example of nontrivial automorphism of $G$.

Therefore, in any case, the group $G$ has a nontrivial automorphism.

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Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.
(Michigan State University, Abstract Algebra Qualifying […]

Isomorphism Criterion of Semidirect Product of Groups
Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

If Squares of Elements in a Group Lie in a Subgroup, then It is a Normal Subgroup
Let $H$ be a subgroup of a group $G$.
Suppose that for each element $x\in G$, we have $x^2\in H$.
Then prove that $H$ is a normal subgroup of $G$.
(Purdue University, Abstract Algebra Qualifying Exam)
Proof.
To show that $H$ is a normal subgroup of […]

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Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

The Order of $ab$ and $ba$ in a Group are the Same
Let $G$ be a finite group. Let $a, b$ be elements of $G$.
Prove that the order of $ab$ is equal to the order of $ba$.
(Of course do not assume that $G$ is an abelian group.)
Proof.
Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
\[(ab)^n=e, […]