Exercise 2.2.2 is
Canonical in "canonical homomorphism" is synonymous to natural. I look for a normal subgroup of $A$. Let $B$ be the set of all $\alpha\in A$ that fix $R$ elementwise. Then $B$ is a subgroup of $A$. Let $\alpha\in A$ and $f\in R$. Then $f(\alpha^{-1}\beta \alpha)=(f\alpha^{-1})\beta \alpha=(f\alpha^{-1})\alpha=f$ because $f\alpha^{-1}\in R$. Therefore $\alpha^{-1}\beta\alpha\in B$ and $B$ is normal in $A$. This is all I could get. Also, suppose $\theta:A\to A/N$ is a natural homomorphism. Then $\alpha\theta=\alpha N$. What is the meaning of $\alpha N\in$ Aut $G$?
“Canonical” here also means that it does not depend on a particular description of the group, but only on its properties (compare for example the isomorphism between a finite dimensional vector space and its dual relative a particular basis, and the isomorphism with the double dual which is “coordinate free”).
Here, if $\alpha\in A$, then $\alpha$ maps $R$ to $R$. Therefore, since $G=F/R$, you can apply $\alpha$ to $G$ by $\alpha(gR) = \alpha(g)R$. You want to show that this is well-defined and gives an automorphism of $G$. This gives you the “canonical map” from $A$ to $\mathrm{Aut}(G)$.