What is this path torsor appearing in this group cohomology formulation of the section conjecture? Let $X$ be a (nice) scheme over $\mathbb{Q}$.  Then there is an exact sequence
$1\rightarrow \pi_1^{et}(\overline{X}, b)\rightarrow \pi_1^{et}(X, b)\rightarrow G_{\mathbb{Q}}\rightarrow 1$.
This defines an outer action $G_{\mathbb{Q}}\rightarrow Out(\pi_1^{et}(\overline{X}, b))$ by taking any lifting to the middle group and using it to conjugate the geometric fundamental group.  Note that a rational point $p\in X(\mathbb{Q})$ provides a section $G_{\mathbb{Q}}\rightarrow \pi_1^{et}(\overline{X}, b)$ by functoriality.  Grothendieck's section conjecture states that every such section comes from a rational point.
In the paper https://arxiv.org/pdf/1910.12755.pdf, Section 4.1 on page 16, the authors state the section conjecture in the following way:
The map $X(\mathbb{Q})\rightarrow H^1(G_{\mathbb{Q}}, \pi_1^{et}(\overline{X}, b))$
is an isomorphism.
My question is: how is the right hand side even defined?  Generally, $H^1(G, M)$ can be identified with crossed homomorphisms/principal ones, when $M$ is a $G$-module.  But in this case $\pi_1^{et}(\overline{X}, b)$ doesn't even carry an action of $G_{\mathbb{Q}}$, only an outer action.  In fact, it takes a section to simply give a well-defined action.  So I'm confused.
EDIT: Soon after writing this, I realized there is another action o $G_{\mathbb{Q}}$ on $\pi_1^{et}(\overline{X}, b)$, simply by letting it act on $\overline{X}$!  However, it is not immediately clear to me how to define a crossed homomorphism under this definition - one natural thing would be to, for a section $s: G_{\overline{K}}\rightarrow\pi_1^{et}(X, b)$ and $\sigma\in G_{\overline{K}}, f\in \pi_1^{et}(\overline{X}, b)$ define
$\sigma \mapsto \sigma[f] = s(\sigma)fs(\sigma)^{-1}f^{-1}$.
Unfortunately I haven't yet been able to check the cocycle condition - maybe it is obvious (or this is just all nonsense) but it is sadly way too late for me to think about it clearly at the moment...
EDIT 2: Upon further reflection about my previous edit I realized that I was being sloppy with basepoints: the "action" I claimed just gives isomorphisms $\sigma: \pi_1^{et}(\overline{X}, b)\rightarrow \pi_1^{et}(\overline{X}, \sigma(b))$.  So I am still confused!
EDIT 3: I have concluded that in fact $b$ is taken to be a rational point!  So there is a well-defined action.  However, I have failed to define the crossed homomorphism.  I would still very much like clarification on this matter.
 A: Thanks to my friends who helped me make sense of this.
So $b$ is taken to be a rational point, so there is a fixed section
$b_*: G_{\mathbb{Q}}\rightarrow \pi_1(X, b)$ which gives an action of $G_{\mathbb{Q}}$ on the geometric fundamental group.  Now given another rational point $s$, we have a map $s_*: G_{\mathbb{Q}}\rightarrow \pi_1(X, s)$.  The claim is that the map
$s\mapsto [\pi_1^{et}(X, b, s)]$
does indeed define a map  $X(\mathbb{Q})\rightarrow H^1(G_{\mathbb{Q}}, \pi_1^{et}(\overline{X}, b))$.
Now of course, I need to define what $\pi_1^{et}(X, b, s)$ is first!  It should be defined as the "paths from $b$ to $s$ up to homotopy", which in our scenario really means isomorphisms of functors $\gamma: F_b\cong F_s$.  By general theory, if $X$ is connected then such an isomorphism exists, and we may once and for all fix a compatible set of such paths.
Now our goal is still to define a crossed homomorphism $G_{\mathbb{Q}}\rightarrow \pi_1^{et}(\overline{X}, b)$ coming from $s$.  We do so by taking inspiration from the principal homomorphisms.  For $f\in \pi_1^{et}(\overline{X}, b)$, these are defined by
$Z(\sigma) = f^{-1}\sigma(f)$.
As a matter of fact, we may extend the Galois action to "paths", by defining $\sigma(\gamma) = \sigma_s\circ\gamma\circ\sigma_b^{-1}$ (here, they are acting on the $\pi_1$-sets the fiber functor gives.  Then we may simply define
$Z(\sigma) = \gamma^{-1}\sigma(\gamma)$.
This is indeed a cocycle for the same reason the principal homomorphisms are!  This answers my question.

Follow-up question for those interested: if I choose a different path $\gamma_2:F_b\cong F_s$, do I get the same cohomology class?

Edit: Both for my original question, and for the follow-up qustion, care must be taken to show that what we get really ends up in the geometric fundamental group, not just $\pi_1^{et}(X, b)$.  (I have resolved this.)
