Group having an element $x$ of order $p$ where $p$ is the smallest prime dividing |G| Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$ and $x\in G$ be an element of order $p$. Let $h\in G,$ and $hxh^{-1}=x^{10}$. Then prove that $p=3$.
If $H=<h>, X=<x>$ then the relation suggests that $H$ acts on $X$ which induces a homomorphism $H$ to $Aut X$ which is of order $p-1$. From that how can one claim that it is trivial homomorphism(i.e. $h\to \phi_{e}$ ?) [THIS IS MY QUESTION], If it happens then $hxh^{-1}=x=x^{10}(given)$ so $x^9 =e$ so $p=3.$
 A: As $hxh^{-1} \in X$, the map $\phi_h : X \rightarrow G$ defined by $\phi_h(x^k)=hx^kh^{-1}$ defines an automorphism of $X$.
In fact, there is a map $\phi : H \rightarrow \mathrm{Aut}(X)$ given by $\phi(h^k)=\phi_h^k$. Now, since $X$ is cyclic of prime order, the automorphisms of $X$ are specified uniquely by sending our particular generator, $x$, to one of $x,x^2,\ldots,x^{p-1}$, so $\mathrm{Aut}(X)$ has order $p-1$.
We want to show that $\phi_h$ is actually the identity element of $\mathrm{Aut}(X)$ i.e. $\phi$ is the trivial homomorphism $H \rightarrow \mathrm{Aut}(X)$. Suppose it isn't. Then $Im(\phi)$ is non-trivial and has order dividing $p-1$. Therefore $|H|$, and so $|G|$ has a non-trivial factor which is less than or equal to $p-1$, contradicting the minimality of $p$. Thus $\phi$ is the trivial homomorphism $H \rightarrow \mathrm{Aut}(X)$, and so $\phi_h$ is the trivial automorphism $X \rightarrow X$, and so $hxh^{-1}=x^{10}=x$, so $x^9=e$ and since $x$ has prime order $p$, $p=3$.
A: Although this is not your approach, here is another solution: Let $n = O(h)$, then since $p$ is the smallest prime dividing $O(G)$, you have
$$
gcd(p-1,n) = 1
$$
Hence, $\exists r,s \in \mathbb{Z}$ such that
$$
r(p-1) +sn = 1 \qquad\qquad(1)
$$
Now note that
$$
x = h^nxh^{-n} = x^{10^n}
$$
So
$$
10^n \equiv 1\pmod{p} \qquad\qquad (2)
$$
Also, since $O(xhx^{-1}) = O(x)$, it follows that $gcd(p,10) = 1$. Hence by Fermat's Little theorem,
$$
10^{p-1} \equiv 1\pmod{p} \qquad\qquad (3)
$$
By (1), (2) and (3), we have
$$
10 \equiv 1\pmod{p}
$$
Hence, $p=3$
