Is the inner automorphism $\phi_g$ the same as $\phi_{g^{-1}}$ In my abstract algebra book, the definition of an inner automorphism is written as:

Let $G$ be a group, and let $a$ belong to $G$. The function $\phi_a$  defined by $\phi_a(x)= axa^{-1}$ for all $x$ in $G$ is called the inner automorphism of $G$ induced by $a$.

However, in Wikipedia, $\phi_a$ is said to be equal to $a^{-1}xa$.
Are they the same? Am I missing anything obvious?
 A: Oh, this is whether you act on the left or the right. It's purely a convention thing.  If you 'act on the right', your conjugate of $x$ by $g$ is $g^{-1}xg$, and group actions on a set are $\Omega\times G\to \Omega$ given by $\omega\cdot g$. If you 'act on the left', your conjugate of $x$ by $g$ is $gxg^{-1}$, and group actions on a set are $G\times \Omega\to \Omega$ given by $g(\omega)$.
Because of Euler's $f(x)$ notation for a function $f$ applied to $x$, rather than $xf$ or $(x)f$, people thought it was a good idea to compose functions from the right side to the left, i.e., opposite to how you read in European languages. But in group theory a lot of people compose from left to right. So homomorphisms $\phi$ are written $g\phi$ rather than $\phi(g)$.
It's a complete pain, and the cause of a lot of confusion and errors.
A: Let $G$ be a group. If $g$ is an element of $G$, we can define two functions:
\begin{align}
&\lambda_g:x\in G\mapsto gxg^{-1}\in G, \\
& \rho_g:x\in G\mapsto g^{-1}xg\in G.
\end{align}
In general, these two functions are different. For example, if $G=S_3$, the symmetric group of degree $3$ and $g=(123)$, then
\begin{align}
& \lambda_g((12)) = (123)(12)(123)^{-1}
= (23)
\\
& \rho_g((12)) = (123)^{-1}(12)(123) = (13).
\end{align}
Sometimes, though, $\lambda_g$ and $\rho_g$ do coincide. An extreme case of this is when the group $G$ is abelian — or, more generally, when $g$ is central in $G$ — for then the two functions are just the identity map of $G$.
Now, for all $g$ we have that $(g^{-1})^{-1}=g$, so that you can immediately check that $$\lambda_g=\rho_{g^{-1}}.$$ This means that the collection of maps $\{\lambda_g:g\in G\}$ coincides with the collection $\{\rho_g:g\in G\}$.
One can wonder why we consider these two maps… Well, let us write $\def\Aut{\operatorname{Aut}}\Aut(G)$ for the group of automorphisms of $G$. The function $$L:g\in G\mapsto \lambda_g\in\Aut(G)$$ is then a morphism of groups. In other words, we have that $L(g)\circ L(h)=L(gh)$ for all choices of $g$ and $h$ in $G$.
What happens with the maps $\rho_g$? Well, the function $$R:g\in G\mapsto\rho_g\in\Aut(G)$$ is not a morphism of groups… One can easily check that $$R(g)\circ R(h)=R(hg),$$ and the right hand side here is in general different from $R(gh)$. One way to «fix» this is to change de domain of the map $R$ into the group $G^{\mathrm{op}}$ opposite to $G$: the map $$R:g\in G^{\mathrm{op}}\to\rho_g\in\Aut(G)$$ is now a perfectly civilized morphism of group.
So what is this? Well, a morphism of groups $L:G\to\Aut(G)$ corresponds to an action of $G$ on itself on the left: for all $g$ and $h$ in $G$ we let $g$ act on $h$ so that $$g\triangleright h=\lambda_g(h).$$ On the other hand, a morphism of groups $R:G^{\mathrm{op}}\to\Aut(G)$ corresponds to an action of $G$ on itself on the right: for all $g$ and $h$ in $G$ we let $g$ act on $h$ so that $$h\triangleleft g=\rho_g(h).$$
People who like their groups acting on the left of things will usually prefer the map $L$, while people who mistakenly prefer their actions on the right usually prefer the map $R$.
