In Lang's "Algebra", p.56 line 5, what does that mean by "Two such isomorphims differ by an inner automorphism"? What's the meaning of "differ"?
It says that: if $u:A\to B$ is an isomorphism, then $Aut(A)$ and $Aut(B)$ are isomorphic under conjugation, namely $w \mapsto\,uwu^{-1}$ is an isomorphism $Aut(A)\to\,Aut(B)$. Two such isomorphims differ by an inner automorphism.
 A: An inner automorphism of a group $G$ is an isomorphism $f:G\to G$ of the form $f(a)= gag^{-1}$ for some $g\in G$. Lang is claiming that if you pick two ring isomorphisms
$$h:A\to B$$
$$g:A\to B$$
then there is an element $a$ of the group $\operatorname{Aut}(B)$ so that if $H$ and $G$ are the induced maps
$$\operatorname{Aut}(A)\to \operatorname{Aut}(B)$$
we have $H(w)=aG(w)a^{-1}$ for each $w\in \operatorname{Aut}(A)$. Put differently, $H=f\circ G$ where $f$ is the inner automorphism defined by $a$, so that $G$ and $H$ differ by $f$. In this case, you can check that $a=hg^{-1}$ is the desired element of $\operatorname{Aut}(B)$. 
A: If we find an isomorphism between two structures, $A$ and $B$, we can regard them as the same.  We have just renamed all the elements of some underlying Platonic structure that we are interested in.  An inner automorphism is an isomorphism that is within the same structure, so it tells us that the elements play the same role within the structure.  A simple example is the cyclic group of order $3$ with elements $\{0,1,2\}$ under addition.  There is an automorphism that exchanges $1$ and $2$, which means there is $$\phi(x)=\begin {cases}0&x=0\\2&x=1\\1&x=2 \end {cases}$$ that makes $\phi(x+y)=\phi(x)+\phi(y)$.  You can check that this works.  The fact that there are no other automorphisms  means that $0$ is special-it is the identity element, but $1$ and $2$ are really the same as being a non-identity element.  
The point of the Lang statement is that all the isomorphisms of the type you are considering are the same.  If we have one isomorphism between $A$ and $B$ (of whatever type you are considering) where $b\in B =\phi(a), a \in A$ and another where $b\in B =\psi(a), a \in A$, there is an inner automorphism in $B$ where $b \leftrightarrow \phi(\psi^{-1}b),$ which says these two isomorphisms are in the same sense the same.  If you have a group of order $3$ that includes $\{a,b,c\}$ and you find an isomorphism $a \leftrightarrow 0,b \leftrightarrow 1,c \leftrightarrow 2$ that is not fundamentally different from the isomorphism $a \leftrightarrow 0,b \leftrightarrow 2,c \leftrightarrow 1$ because the inner automorphism says they are the same.
