Suppose $G$ is a group. For each $g\in G$ we define the function $\phi_g(x) : G\to G$ as $\phi(x) = g^{-1}xg$. Prove that $\phi_g(x)$ is an automorphism of $G$.
This is what I did ( But I don't know if it's right ):
The only possibility for me that $\phi$ being an automorphism was if it's an identity function . So:
$$\begin{align} \phi_g(g)& = g^{-1}gg\\ \phi_g(g) &= (g^{-1}g)g\\ \phi_g(g) &= eg\\ \phi_g(g)& = g. \end{align}$$
But at the same time this doesn't seem quite right for me...
Fix $g\in G$. The function $\phi:G\to G$ is injective and well-defined because for all $x,y$ we have
$$\begin{align} x=y&\iff g^{-1}x=g^{-1}y\\ &\iff g^{-1}xg=g^{-1}yg\\ &\iff \phi(x)=\phi(y). \end{align}$$
It is surjective because, for any $h\in G$, we have $h=g^{-1}(ghg^{-1})g=\phi(ghg^{-1})$ while $ghg^{-1}\in G$.
It is a homomorphism, too, since for any $a,b\in G$, we have
$$\begin{align} \phi(ab)&=g^{-1}(ab)g\\ &=g^{-1}a(gg^{-1})bg\\ &=(g^{-1}ag)(g^{-1}bg)\\ &=\phi(a)\phi(b). \end{align}$$
Hence $\phi$ is an isomorphism from $G$ to $G$; that is, $\phi$ is an automorphism.