Let $B_n$ be the braid group with $n$ strands and let $F_n$ be the free group of rank $n$ generated by $x_1,\ldots,x_n$. The classical Artin Representation Theorem reads:
If an automorphism of $F_n$ is an isomorphism from $F_n$ to itself, then what is a right automorphism?
Automorphisms are special cases of group actions - the automorphism group acts on the underlying set of some first group. In particular this is a left group action. If $A$ is a group and we have a set $X$, recall a left action of $A$ on $X$ is a map $A\times X\to X$ denoted $(a,x)\mapsto ax$ which satisfies the "associativity" relation $a(bx)=(ab)x$ for all $a,b\in A$ and $x\in X$. Similarly then we can define a right group action as a map $X\times A\to X$ with $(xa)b=x(ab)$ for all $a,b,x$. Even though functions are normally written on the left (e.g. $f(x)$) sometimes they can be on the right side instead (e.g. $(x)f$). It is somewhat unusual though. The right automorphism group is then the set of all right functions of $G$ which are automorphisms of $G$.
The point of right actions is to keep track of how the actions compose together in a consistent and correct manner; sometimes naturally occurring actions are not left actions. For instance we know that $S_n$ acts on $\{1,\cdots,n\}$ and so it acts on $\{(x_1,\cdots,x_n):x_i\in X\}$ for any set $X$ by permuting the coordinates. If $\sigma$ is to put $x_i$ into the $\sigma(i)$-coordinate though, that means $(\sigma x)_{\sigma(i)}=x_i$, or in other words (via the substitution of $\sigma^{-1}(i)$ for $i$) $\sigma(x_1,\cdots,x_n)=(x_{\sigma^{-1}(1)},\cdots,x_{\sigma^{-1}(n)})$. The fact that $(x_1,\cdots,x_n)\sigma=(x_{\sigma(1)},\cdots,x_{\sigma(n)})$ is a right action, not a left action, often trips many people up, even authors of lecture notes in my experience.