A question on finite group 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 $p=3$ ?  
 A: 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 $o(x)=p$ 
$\Rightarrow p=3$
A: 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.
