How to show that naturally isomorphic functors preserve the same limits? I've seen a number of times versions of the claim that if $F$ is naturally isomorphic to $G$, then if $F$ preserves limits of shape $\mathbf{J}$ so does $G$. (Sometimes with "It is easy to show that ...."!)
But my attempts to prove it have got into a tangle, so I must be missing some simple idea. I'd be very grateful then for (a sketch of) a proof!
 A: D_S gives the headline idea. Perhaps the key thing to see is that -- in the terms of the proof below -- each face of the prism, not just the bottom face, is a naturality square. That gives you a lot to play with! So here's the proof spelt out rather more. And I've slightly edited things [from a version of notes of my own!] to fit the question:

A: Let $F, G$ be functors from $\mathscr C$ to $\mathscr D$.  On one piece of paper, you have a commutative diagram of objects and morphisms in $\mathscr C$.  On a second piece of paper, you trace out the identical commutative diagram in $\mathscr D$ where you apply the functor $F$ to the first diagram.  On a third piece of paper, you trace out the identical commutative diagram in $\mathscr D$ where you apply the functor $G$ to the first diagram.  
Line up the second and third pieces of paper to be parallel and facing each other about an inch apart, and then draw a lines (about an inch long) between the corresponding objects.  Then every such line is an isomorphism, and you get a 3-D commutative diagram composed of two layers, both essentially the same, linked by compatible isomorphisms.
