Let $G$ be a group and $g$ be one fixed element of $G$. Show that the map $i_g(x)=gxg'$ for $x \epsilon G$ is an isomorphism of G with itself.
One of the things that I couldn't understand is under what binary operation the expression $gxg'$ is. Is it multiplication? If multiplication why isn't it given under first place. If it's a binary operation $*$ that we have no knowledge of, then why is this written in this way?
Let us prove that $i_g: G \to G: x \mapsto gxg^{-1}$ is a group isomorphism. (I assumed you meant with $g'$ the inverse of $g$, so I wrote $g^{-1}$ instead of $g'$). Note that we do not have to write down the binary operation, since we are working in $G$ the entire time.
It is clear that $i_g(G)$ is a subset of $G$. Let us prove that $i_g$ is a group homomorphism: let $x, y \in G$. Then we have that \begin{align} i_g(xy) &= gxyg^{-1}\\ &= gx(g^{-1}g)yg^{-1}\\ &= (gxg^{-1})(gyg^{-1})\\ &= i_g(x)i_g(y) \end{align} showing this is a group morphism.
Let us now determine the kernel of this map: If $x \in \ker(f)$, then we have that $gxg^{-1} = e_G$ ($e_G$ the neutral element in $G$) and therefore $x = e_G$ (do you see why?).
Moreover, $i_g$ is surjective: let $y \in G$ and consider the element $g^{-1}yg \in G$. We have that $i_g(g^{-1}yg) = g(g^{-1}yg)g^{-1} = y$.
This proves that $i_g$ is a bijection and a morphism, hence an isomorphism. Such an isomorphism (from a group to itself) is called an automorphism.