Given some index category $\mathcal{I}$ and a complete category $\mathcal{C}$, we consider the diagonal functor $\Delta : \mathcal{C} \to [\mathcal{I}, \mathcal{C}]$ to be the functor that sends all objects $C$ of $\mathcal{C}$ to the constant diagram $\Delta(C)$ that maps each $I$ to the fixed object $C$? In particular, I want to show that the assignment that maps each diagram $D$ to its apex $X_D$ of its limit cone is itself a functor, say, '$\text{lim} : [\mathcal{I}, \mathcal{C}] \to \mathcal{C}$'. (Remember that we take our category $\mathcal{C}$ to be complete, or at least, to have all limits of shape $\mathcal{I}$.) In doing so, I first got confused, but then that confusion led to a possible insight. The confusion arose when I considered the action of $\Delta$ on morphisms (which in our case happen to be natural transformations $\mu : D \Rightarrow D'$). I wrote down $\Delta_{\mu}$, but soon realised that things were not well-defined. Then I came up with the idea to consider the diagonal functor 'one level up', in the following sense: $\Delta^\top : [\mathcal{I}, \mathcal{C}] \to [\mathcal{I}, [\mathcal{I}, \mathcal{C}]]$. I hoped this would prove useful to avoid lengthy verifications for showing functoriality (i.e., preservation of identities and morphisms) of 'lim', so I hoped that using this 'new' diagonal functor ($\Delta^\top$), of which we know that it is both itself a functor, but so are its images on the objects of $[\mathcal{I}, \mathcal{C}]$. Would this indeed be useful or just replace tedious verifications of functoriality of lim on morphisms to tedious verifications of well-definedness of $\Delta^\top$ on objects?
Concretely, can I define the morphism part of lim in terms of $\Delta^\top$ in such a way that I need not spell out the universal property of the terminal cone?
(Note: I need a very 'entry-level' proof, i.e., absent notions of Yoneda or adjunctions.)
$\newcommand{\C}{\mathsf{C}}$$\Delta$'s action on morphisms does not go for $\Delta(\mu)$ where $\mu$ is a natural transformation. $\Delta$ has domain $\C$ so the relevant morphisms are arrows within $\C$. As commented, there is only one obvious way that works in general to define $\Delta(f)$ for some $f:\varsigma\to\varsigma'$ in $\C$; it is to let it be the natural transformation with every component equal to $f$, $\Delta(\varsigma)\implies\Delta(\varsigma')$.
Things are well-defined here. You probably encountered issues because you mixed up the domain with the codomain; $\Delta(\mu)$ does not make sense. It is the $\lim$ functor which has to act on natural transformations and map these to arrows in $\C$, not $\Delta$ itself. The preservation of identities and composition by $\lim$ is quite easy to show and does not require lengthy verification. Use your universal properties (I'll elaborate on that if requested).
$\Delta$ and $\Delta^{\top}$ are both well-defined and this requires no verification at all; it is certainly not tedious. I really think you've just conflated the domain of $\Delta$ with that of $\lim$ and have become confused because of it.