How to eliminate other possibilities of $p$? Let $G$  be a finite group, $p$ be  the smallest prime divisor of $|G|$, and $x \in  G $ an element of order p. Suppose
$h \in  G $ is such that $hxh ^{−1} = x ^{10} $. Show that $p=3$
My try:
Since $o(x)=p$ and $hxh^{-1}$ and $x$ being conjugates have the same order we have $o(hxh^{-1})=o(x^{10})=p$
Also $o(x^{10})=\dfrac{p}{\gcd(10,p)}=p$
Then $\gcd(10,p)=1$ then $10^{p-1}\equiv 1(\mod p)$
I can see that $p=3$ satisfies the equation .Also $p\neq 2,5$.How to eliminate other possibilities of $p$?
 A: Let $H=<h>$ and $X=<x>$, and $Aut(X)$ is the group of automorphisms of $X$
because $x^{10}=hxh^{-1}$ translates the fact that $H$ acts on $X$ by conjugation, in other words, we have a homomorphism $\varphi : H → Aut(X)$ for any element $h$ and any element $t$ we have $\varphi(h)(x)=hth^{-1}$.
This latter group has order $p−1$ and the first group has order with all prime divisors $\geq p$ whence $\varphi$ is the trivial homomorphism. This means that $hxh^{-1} = x$. But we are given that $hxh^{-1} = x^{10}$. It follows that $x^9$ is the identity element, so $p = 3$.

Another proof (simplification of the notions used in the first proof)
One can see that $x^{10}=hxh^{-1}$ implies that $x^{10^q}=h^{q}xh^{-q}$ using Fermat's little theorem combined with $q=p-1$ we get $x=h^{p-1}xh^{-(p-1)}$ . Let's now take $q$ as the order of $h$, as we know the order of $h$ is coprime with $p-1$ hence there exist some $u$ such that $u(p-1)\equiv 1 \mod q$ and then:
$$x=h^{p-1}xh^{-(p-1)}=h^{2(p-1)}xh^{-2(p-1)}=h^{u(p-1)}xh^{-u(p-1)}=hxh^{-1} $$
which gives you the result.
