What is the difference between Path $\infty$-groupoid and Smooth Fundamental $\infty$-groupoid of a smooth space?

Question

A couple of days back I asked a question Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space? in MO about the existence of a possible Smooth/Geometric version of Homotopy Hypothesis using the notion of Path $\infty$-groupoid of a smooth space.

After a discussion in the comments section with @David Roberts I got a feeling (but not completely convinced) that though Path 1-groupoid and smooth fundamental 1-groupoid of a smooth space are quite different objects but "if we move upto infinity level" and present them as Kan Complexes then they are becoming the same object.

3 months back I asked the following MO question What is the geometric realization of the the nerve of a fundamental groupoid of a space?.

From the discussions in

1. Is there a Geometric/Smooth version of Homotopy Hypothesis using the path $\infty$-Groupoid of a Smooth Space?

2. What is the geometric realization of the the nerve of a fundamental groupoid of a space?

now I have the following Questions/Doubts:

We know that the construction of Smooth Fundamental 1-Groupoid and Path 1-Groupoid of a smooth space induce natural functors $Man \rightarrow Groupoids$. Now from the discussion in What is the geometric realization of the the nerve of a fundamental groupoid of a space? I expect that $|N \circ \pi_{\leq 1}(X)|$ contains all informations of the 1st Homotopy groups of the smooth space $X$ where $N$ is the Nerve functor, $\pi_{\leq 1}$ is the Smooth Fundamental 1-Groupoid functor and $|-|$ is the Geometric realization functor. Now we can repeat the same procedure with Path 1-Groupoid functor $\pi'_{\leq 1}: Man \rightarrow Groupoids$.

My questions are the following:

1. Is $|N \circ \pi_{\leq 1}(X)|= |N \circ \pi'_{\leq 1}(X)|$? (where "$=$" is in an appropriate sense)

2. Is there a way to present a Path $\infty$-groupoid of a smooth space such that it is different from Smooth Fundamental $\infty$-groupoid of the space? (So that it matches our intuition for $n=1$ case)

(By "$n$" I mean "Groupoids in the level 1").

Answer 1

I can only answer your first question, and the answer is no. Take for example $X=\mathbb{R}^2$, so that the fundamental groupoid is trivial, but the path groupoid contains distinct arrows represented by circles of every positive radius passing through a fixed basepoint (and many many more besides). This is ignoring all questions of topology or smooth structure on the set of arrows, which I think is your intent. And so the geometric realisations of the nerves of these cannot even be weakly homotopy equivalent, as one is contractible and one has fundamental group that is not even finitely generated.