Do equivalent index categories give isomorphic colimits?

Question

Imagine functors $F \colon J \to \mathscr{C}$, $G \colon I \to J$, where $G$ is an equivalence of categories. I want to know if "$\underset{i \in I}{\text{colim}} \big(F \circ G\big)(i) = \underset{j \in J}{\text{colim}} F(j)$".

To make this question a bit more precise: Colimits are defined as $\textit{limiting cones}$ in Saunders Mac Lanes "Categories for the working Mathematician".

A $\textit{cone with base F }$ is a pair $(c, u)$ consisting of an object $c \in \mathscr{C}$ and a natural transformation $u \colon F \Rightarrow \Delta_J(c) $. Here $\Delta_J(c) \colon J \to \mathscr{C}$ is the functor that maps $j \mapsto c$ and $(f \colon j \to j') \mapsto \text{id}_c$.

A $\textit{limiting cone with base F }$ is a cone $(c,u )$ such that for all other cones $(d,v)$ with base $F$ there is a unique morphism $f \colon c \to d$ such that $v = \Delta_J(f) \cdot u$. Here $\Delta_J(f) \colon \Delta_J(c) \Rightarrow \Delta_J(d)$ is the natural transformation with $\big(\Delta_J(f)\big)_j = f$ for all $j \in J$.

Now imagine a limiting cone $(c,u)$ with base $F \circ G$ and a limiting cone $(d,v)$ with base $F$. My Questions is: Is there a natural transformation $u' \colon F \Rightarrow \Delta_J(c)$ such that $(c, u')$ is a limiting cone with base $F$?

Answer 1

Yes. The equivalence is enough to tell you that all diagrams in $I$ have isomorphic replacements by diagrams in $J$, these replacements being coherent through naturality. Don't forget that a natural equivalence $G: I \to J$ also has an inverse natural equivalence, so the two diagram categories really do carry the same data.

Said differently, it is easy to prove that an equivalence $G: I \to J$ of small categories induces an obvious equivalence $$ G^* : [J, \mathcal{C}] \to [I, \mathcal{C}] $$ of functor categories, so colimits coincide.