**1.**  Here is what Tate says in his account of the General Reciprocity Law in the AMS volume on Hilbert's problems :

> With this work of Takagi the theory of
> abelian extensions --- "class field
> theory" --- seemed in some sense
> complete, yet there was still no
> general reciprocity law.  It remained
> for Artin to crown the edifice with
> such a theorem.  He conjectured in
> 1923 and proved in 1927 that there is
> a *natural* isomorphism $$
> C_K/N_{L|K}C_L\buildrel\sim\over\to\operatorname{Gal}(L|K)
> $$ which is characterised by the fact
> that...

And a little later :

> How did Artin guess his reciprocity
> law ?  He was not looking for it, not
> trying to solve a Hilbert problem. 
> Neither was he, as would seem so
> natural to us today, seeking a
> canonical isomorphism, to make
> Takagi's theory more functorial.  He was led to the law by trying to show...

Read him.


 
**2.** Here is a toy example --- not unrelated to class field theory --- of how a bijection can be more natural than others.  Let $p$ be a prime number and let $K$ be finite extension of $\mathbb{Q}_p$ containing a primitive $p$-th root of $1$.  There are only finitely many degree-$p$ cyclic extensions $L|K$, and there are only finitely many vectorial lines in the $\mathbb{F}_p$-space $K^\times/K^{\times p}$.  In fact the two sets have the same number of elements, but the only natural bijection is 
$$
L\mapsto\operatorname{Ker}(K^\times/K^{\times p}\to L^\times/L^{\times p}),
$$
of which the reciprocal bijections can be written $D\mapsto K(\root p\of D)$.

It follows that the number of degree-$p$ cyclic extensions $L|K$ is the same as the number of hyperplanes in $K^\times/K^{\times p}$.  But is there a natural bijection between these two sets ?  You will agree that $L\mapsto N_{L|K}(L^\times)/K^{\times p}$ is as natural a bijection as there can be.

One last point : Given a hyperplane $H\subset K^\times/K^{\times p}$, how do you recover the degree-$p$ cyclic extension $L|K$ such that $H=N_{L|K}(L^\times)/K^{\times p}$ ?  Answer : use the *natural* reciprocity isomorphism $K^\times/K^{\times p}\to\operatorname{Gal}(M|K)$, where $M|K$ is the maximal elementary abelian $p$-extension, to identify $H$ with a subgroup of $\operatorname{Gal}(M|K)$, and take $L=M^H$.