Understanding the determinant example of a natural transformation. This question is about intuition. In the example below, how do we understand the natural transformation $\det$ as a relationship between the two functors given rather than as a relationship between ring homomorphisms and the determinant.
I think this question differs from this similar one because it's about leveraging an understanding of the determinant map to understand natural transformations, rather than proving that the determinant map is an n.t. This question, similarly, seems to be about the proof rather than the intuition explicitly.
So, a natural transformation is an elementary notion in category theory. That said, I find them very difficult to understand and have been struggling with them off and on for a few years. The technical definition of $\eta : F \Rightarrow G$ and the meaning of a naturality square is pretty straightforward.
$$ \require{AMScd}
\begin{CD}
F(X) @>\eta(X)>> G(X)\\
@V F(f) VV @VV G(f) V\\
F(Y) @>\eta(Y)>> G(Y)
\end{CD} $$
Let $\varphi$ be the implied arrow. The above diagram says:
$$ \eta(Y) \circ F(f) = \varphi = G(f) \circ \eta(X) $$
What I do find difficult to understand is the motivation behind this definition ... and how to take a concrete natural transformation and "scrape away the detail" to get the real definition.
For example, you can get the idea of a ring by taking $\mathbb{Z}$ and scraping away some of the detail to get an algebraic theory.
This question has an interesting example of a natural transformation.
I want to focus on this example because it is concrete and there's a familiar concrete algorithm associated with the determinant.
$$ \require{AMScd}
\begin{CD}
\mathrm{GL}_nK @>\det_K>> K\text{*}\\
@V\mathrm{GL}_nfVV @VVf\text{*}V\\
\mathrm{GL}_nK^\prime @>\det_{K^\prime}>> K^\prime\text{*}
\end{CD} $$
Let $\varphi : \text{GL}_n(K) \to K^*$, not shown above, be the implied arrow in the commutative diagram.
I think the claim is $ \det : \text{GL}_n \Rightarrow (+)^* $ is a natural transformation where $(+)^*$ is a covariant functor.

*

*I'm pretty sure this whole diagram takes place in the category $\mathsf{Group}$. (I don't know whether we lose anything if we make a deliberately strange choice and insist that it lives in $\mathsf{Monoid}$, though.)

*$K$ is a commutative ring. $K'$ is also a commutative ring.

*$K^*$ is the group of units of $K$ (sometimes written $K^\times$). $K^*$ and $K'^*$ are Abelian groups because the rings they are drawn from are commutative.

*$\det_K$ is the determinant map. For finite dimensional $K$ we can use the Leibniz formula to compute the determinant explicitly for any finite-dimensional matrix with coefficients taken from a commutative ring.

*$f : K \to K'$ is a ring homomorphism.

*I buy that $\text{GL}_n$ is a functor.

*I buy that $(+)^*$ is a functor.

So, the commutative diagram is saying that our implied arrow, $\varphi$, the map taken by computing the determinant and then applying our ring homomorphism could also have been done in the other order.
We might express this idea without involving category theory in the following way.
$$ \text{det}(f(M)) = f(\text{det}(M)) \;\; \text{where $M$ is in $\text{GL}_n(K)$} $$
And then we carefully explain that $f(M)$ means applying $f$ coordinate-wise to the entries of $M$.
This makes sense as a statement about $\det$ and $f$ ... and we can show that the elementwise extension of $f$ is a conservative extension of $f$ (and oddly reminiscent of APL).
However, I'm struggling to see how to change my perspective to view it as some kind of relationship between $\text{GL}_n$ and $(+)^*$ ... because $(+)^*$, at an intuitive level, does so little. $(+)^*$ just takes away my non-invertible elements and takes away my $+$ symbol.
 A: One way to interpret natural transformations is as operations derived from the structure of an object not the particular object itself. In your example, the determinant formula as a sum over permutations of signed products of variables makes sense whenever the variables range over elements of a ring. However, the formula does not depend on which ring your are working with. It is derived from only the ring structure (associative and commutative addition, subtraction, associative and commutative multiplication).
Now a ring homomorphism preserves the ring structure: doing addition or multiplication followed by a ring homomorphism is the same as doing the ring homomorphism first and then the addition or multiplication. This means that it should also preserve, in the same sense, any operation derived from the ring structure alone, e.g. the determinant.
More generally, you can interpret morphisms of a category as "structure"-preserving functions between objects. Functors would then be constructions which depend only on the structure in the sense that "structure"-preserving functions also act on the constructions (e.g. apply ring homomrphism element-wise to a matrix). The natural transformations are then transformations between constructions that depend only on the structure of the object on which the constructions are performed, not the object itself. For example, the determinant as an operation converting square matrices to rank $1$ matrices.
