A short answer would be: $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts faithfully on the profinite fundamental groupoïd of the operad of little discs.

If $X$ is an algebraic variety over $\mathbb{Q}$ we have an exact sequence
$$
1 \to\pi_1(X\otimes \overline{\mathbb{Q}},p) \to
\pi_1(X,p) \to Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to 1
$$
Here $\pi_1(X\otimes \overline{\mathbb{Q}};p)$ is canonically identified with the profinite completion of the usual topological fundamental group $\pi_1(X(\mathbb{C}),p)$.
If the basepoint is defined over $\mathbb{Q}$, this split and we have an action
$$
Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to Aut(\widehat{\pi}_1(X(\mathbb{C}),p)).
$$

The (profinite completion of the) fundamental groupoïds of the $C_2(n)$ inherit the operad structure. The trick is that all of it can be defined over $\mathbb{Q}$ as $C_2(n)$ is homotopy equivalent to the configuration space of points on the affine line $F(\mathbb{A}^1_{\mathbb{Q}},n)(\mathbb{C})$. One has to define rational "tangential base points" and check that the operad structure on the fundamental groupoïds is also defined over $\mathbb{Q}$. The resulting operad is described here. One can explicitly compute its automorphism group. This is the Grothendieck-Teichmuller group $\widehat{GT}$.

As everything is defined over $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ operates on the whole operad. So we have a morphism
$$
Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to \widehat{GT}
$$
It follows from a theorem of Belyi that it is injective.