Q: If the limits of two functors are isomorphic in a category, do we know anything about the functors? Say $F:I\to \mathcal{C}$ and $G:J\to \mathcal{C}$ are diagrams in $\mathcal{C}$ and suppose that, in $\mathcal{C}$, $$\varprojlim F\cong \varprojlim G.$$
Can we say anything about $F$ and $G$? Are they, for example, isomorphic?

A: I do it for colimits, you can easily dualize.
Let $\cal I$ be a small category. For each $A\in \cal I$ the functor $\hom(A,-) \colon {\cal I}\to \bf Set$ has the terminal set as colimit, with the unique possible cocone, and yet $\hom(A,-)\cong \hom(B,-)$ iff $A\cong B$; take a category with at least two non isomorphic objects. :-)

A: Even if $I=J$, $F$ and $G$ still can do rather different things.
Take $I=J=3=\{0,1,2\}$, i.e. the 3-element discrete category with no morphisms (except identities). Take $\mathcal{C}$ to be the "diamond" poset/lattice:
   T
  /|\ 
 / | \ 
x  y  z
 \ | /
  \|/
   B

Since we have a lattice, any (finite) limit exists and is simply the infimum.
Define $F$ and $G$ by:
$$
\begin{array}{lll}
F(0) &=& x \\
F(1) &=& y \\
F(2) &=& z
\end{array}
\quad\quad\quad
\begin{array}{lll}
G(0) &=& y \\
G(1) &=& z \\
F(2) &=& y
\end{array}
$$
So we have
$$
\lim F = \inf\{x,y,z\} = B = \inf\{y,z\} = \lim G
$$
even though $F$ and $G$ are very different:


*

*$F(0) \not\cong G(0)$, hence $F\not\cong G$

*$F(i) \not\cong G(i)$, for all $i\in I$

*$F$ does not identify non-isomorphic, but $G$ does.

