STATEMENT: Let $A,B$ beobjects of a category $\mathcal{A}$. Let Iso$(A,B)$ be the set of isomorphisms of Awith B. Then the group Aut$(B)$ opoerates on Iso(A,B) by composition; namely, if $u\in$Iso$(A,B)$ and $v\in$Aut$(B)$, then $(v,u)\mapsto v\circ u$ gives the operations. If $u_0$ is one elementof Iso$(A,B)$, then the orbit of $u_0$ is all of Iso$(A,B)$, so $v\mapsto v\circ u_0$ is a bijection Aut$(B)\mapsto$Iso$(A,B)$. The inverse mapping is given by $u\mapsto u_0u_0^{-1}$.
QUESTION: I don't understand how Lang gets the inverse mapping at then end.
I am sure the inverse map $\alpha: \operatorname{Iso}(A,B)\to\operatorname{Aut}(B)$ is $u\mapsto u\circ u_0^{-1}$. Any $u\in\operatorname{Iso}(A,B)$ can be written as $u=v\circ u_0$ for some $v\in\operatorname{Aut}(B)$, as you already pointed out. Hence, $\alpha(u)=u\circ u_0^{-1}=v\circ u_0\circ u_0^{-1}=v$. This is clearly an inverse to $v\mapsto v\circ u_0$.