While working on a homework assignment, I wondered if there is a notion of natural transformation (and isomorphism) of functions as there is in category theory for functors.
My idea was to declare two functions $f:A\to B$ and $g:C\to D$ equivalent iff
$A\cong C$ via $\varphi_1$,
$B\cong D$ via $\varphi_2$
and $f = \varphi_2^{-1} \circ g \circ \varphi_1$.
It's not necessary to read this, but if you want to see my motivation, I've included it here...
My homework assignment asked me to show that $R[x]$ is flat for a commutative ring $R$. So, I attempted to do this by demonstrating that the functor $T_{R[x]}$ is exact. To do this, I would have to show that the sequence
$$0\to T_{R[x]}(M_1)\to^{T_{R[x]}(f)} T_{R[x]}(M_2)\to^{T_{R[x]}(g)} T_{R[x]}(M_3)\to 0$$ is exact if $$0\to M_1\to^f M_2\to^g M_3\to 0$$ is exact.
But the first sequence is just
$$0\to M_1\otimes_R R[x]\to^{f\otimes_R1} M_2\otimes_R R[x]\to^{g\otimes_R1} M_3\otimes_R R[x]\to 0$$
So, I recognized that $R[x]\cong\oplus_{\mathbb{N}}R$ and identified the sequence
$$0\to M_1\otimes_R R[x]\to^{f\otimes_R1} M_2\otimes_R R[x]\to^{g\otimes_R1} M_3\otimes_R R[x]\to 0$$ with the sequence
$$0\to M_1\otimes_R (\oplus_{\mathbb{N}}R)\to^{f\otimes_R(\oplus_{\mathbb{N}}1)} M_2\otimes_R (\oplus_{\mathbb{N}}R)\to^{g\otimes_R(\oplus_{\mathbb{N}}1)} M_3\otimes_R (\oplus_{\mathbb{N}}R)\to 0$$
where $\oplus_{\mathbb{N}}1$ is defined on $\oplus_{\mathbb{N}}R$ via $(r_1,r_2,...)\mapsto (1(r_1),1(r_2),...)=(r_1,r_2,...)$ (just the identity on $\oplus_{\mathbb{N}}R$)
and I want to identify the functions $f\otimes_R1:M_1\otimes_R R[x]\to M_2\otimes_R R[x]$ and $f\otimes_R(\oplus_{\mathbb{N}}1):M_1\otimes_R(\oplus_{\mathbb{N}}R)\to M_2\otimes_R(\oplus_{\mathbb{N}}R)$ using the identification proposed above. I would want to do this for the g's as well...
If this makes sense, I can continue to identify tensors and functions between tensors until I've shown the function I ultimately (there will be several more identifications until I get the sequence I want) identify $f\otimes_R1$ with is injective, thus proving the $f$ is injective implies $f\otimes_R1$ is injective and so proving that $T_{R[x]}$ is exact (and thus, $R[x]$ is flat).
Your definition of morphism between two functions is precisely what we obtain as morphisms on the category of arrows of $\mathbf{Set}$: https://ncatlab.org/nlab/show/arrow+category
The way I think of it: Given a function $f:A\to B$, we draw a graph with the points of $A$ and $B$ (seen as disjoint sets), and draw an arrow from each point $a\in A$ to $f(a)\in B$. Then we have a graph and a certain ordered partition $(A,B)$ of the vertex set. Do the same thing for $g$. A morphism between the two functions is the same thing as a morphism between these graphs which respects the partition.