A question on finite group

Question

Let $G$ be a finite group and $p$ be the smallest prime divisor of $|G|$ , let $x \in G$ be such that $o(x)=p$ , and suppose for some $h\in G $ , $hxh^{-1}=x^{10}$ , then is it true that $p=3$ ?

Answer 1

Yes.
Since $o(x)=p,\ x\neq e. $
Suppose $p=2.$
Then $o(x)=p=2 \Rightarrow x^{2}=e.$
And hence $hxh^{-1}=x^{10}=(x^{2})^{5}=e^{5}=e \Rightarrow x=h^{-1}h=e;$ contradiction.
Therefore $p \neq 2 \Rightarrow 2$ is NOT the smallest prime dividing $|{G}| \Rightarrow 2 \nmid |G| \Rightarrow |G|$ is odd.

Consider the action of $G$ on itself by conjugation. Let $C(x)$ be the conjugacy class of $x$. Observe that $x$ and $x^{10}$ are conjugates (Since $hxh^{-1}=x^{10}$). What can we say about $y$ and $y^{10}$ for $y \in C(x)?$
Let $y \in C(x)$. Then $y=gxg^{-1}$ for some $g \in G.$ Then,
$y^{10}=(gxg^{-1})^{10}=gx^{10}g^{-1}=g(hxh^{-1})g^{-1} = ghxh^{-1}g^{-1}=(gh)x(gh)^{-1} \Rightarrow y^{10} \in C(x).$

That is, we have shown that $y \in C(x) \Rightarrow y^{10} \in C(x).$
If $y \neq y^{10} \ \ \forall y \in C(x),$ then elements of $C(x)$ can be partitioned as $\big \{y, y^{10} \big \},$ which will imply that $|C(x)|$ is even $\Rightarrow |G|$ is even, which is a contradiction.

Therefore $\exists z \in C(x)$ such that $z=z^{10}.$ Now, $z \in C(x) \Rightarrow z=g_{1}xg_{1}^{-1}$ for some $g_{1} \in G.$
Thus, $z=z^{10} \Rightarrow g_{1}xg_{1}^{-1} = (g_{1}xg_{1}^{-1})^{10} = g_{1}x^{10}g_{1}^{-1} $
$ \Rightarrow g_{1}xg_{1}^{-1} = g_{1}x^{10}g_{1}^{-1}$
$\Rightarrow x=x^{10}$
$\Rightarrow e=x^{9}$
$\Rightarrow p \mid 9,$ since $o(x)=p$
$\Rightarrow p=3$

Answer 2

Since the order of $x \in G $ equals that of $h x h^{-1},\forall h \in G$,and then we have $$ p = | h x h ^{-1} | = | x^{10} | = \frac{p}{\gcd(p,10)} $$ It follows from the above equations $$ \gcd(p,10 ) = 1 $$ With $10= 2\times 5$,we must have $$ p \neq 2 $$ and $$ p \neq 5 $$ and in all primes,there is a prime $3$ which is the smallest.