Yeah that is more or less what I expected. Thanks, Tom.

Thanks for the clarification. To take the last statement further, if I demand that the embeddings preserve a framing of $V$ do I get a homotopy equivalence to $F_k$ with unlabeled points, and so the highly connected as $n\to\infty$ result?

No not proper, just smooth. I think the answer below is what I need.

Smooth embeddings as manifolds. Proper I'm not sure...

Yes as long as one sticks to strict 2-categories and strict functors, the details in proving the lax transformation identity are not too terrible using the obvious definition of lax wedge/lax end. I'll take that as evidence that obvious is right in this case. Still, a reference would be nice.

I guess there's an obvious candidate for what a "lax wedge" $c\Rightarrow F$ should be. Maybe the rest is straightforward too...

Thanks, Finn. I'm still holding out hope for a reference in English, but all of the above is helpful.

@David: The Gray tensor product has more data than the cartesian product, which makes the problem worse, i.e. there are "more" p.functors $J\otimes [1]\to\mathcal{C}$ than p.functors $J\times [1]\to\mathcal{C}$, because $J\times [1]$ embeds in $J\otimes [1]$.

@Martin: I'm not sure what you mean by the middle paragraph, but assuming you meant a(j)o F(g)=>G(g)o a(j), you actually do have this data on both sides. Clearly it is part of the natural transformation data, and for a p.functor $h:J\times [1]\to C$, this isomorphism is produced by applying the compositors of $h$ to the equal factorizations [ (j'\to j,\text{id}_1)\circ (\text{id}_j, 0\to 1) = (\text{id}_{j'},0\to 1)\circ (j'\to j, \text{id}_0) ] in the middle we have $h(j'\to j,0\to 1)$, which is the datum that has no counterpart on the p.natural transformation side

Thanks, Tim, can you go into more detail please?

@JC define the Bring radical to be the polynomial itself.