Rough paths can be thought of as taking values in a Lie group embedded in the tensor algebra of $\Bbb R^d$. See page 17/section 2.3. Lie groups represent the continuous symmetries of some object. That is, elements of the Lie group act on some other object in a way that preserves symmetry.
I am curious, since rough paths are elements of a Lie group, what exactly are they acting on? What symmetry do rough paths preserve? What do those actions look like?
Though I like the Arnoldian spirit of the question ("a group is not some set with a forgettable system of axioms but something which acts on a space"), I think the comments given above are already spot on: Rough paths are paths of a certain regularity (often one takes finite $p$-Variation or equivalently $1/p$-Hölder type paths) with values in certain Lie groups, e.g. taking paths with finite $p$-Variation in $\mathbb{R}^d$,Chen's theorem asserts that the signature of such a path takes values (in a suitable sense) in the free $\lfloor p\rfloor$-step nilpotent Lie group $G^{\lfloor p\rfloor}(\mathbb{R}^d)$ (a so called Carnot-Caratheodory group).
However, due to the seminal results of Lyons, one can always (uniquely) lift these paths to paths with values in $G^{N}(\mathbb{R}^d)$ for $N \geq \lfloor p \rfloor$ and this process (the so called Lyons lift) is continuous in the $p$-variation topologies. So rough paths (of a prescribed regularity) can be seen to have values in many Lie groups. Indeed one can lift these rough paths even to the projective limit of the nilpotent groups, which happens to be an infinite-dimensional Lie group (and can be identified with the character group of a certain Hopf algebra).
Moreover, the paths (of a given regularity) form a group under concatenation of paths. The signature map (taking a path to its signature) identifies the paths of a given regularity with a subgroup of the projective limit $G = \lim_{p \rightarrow \infty} G^{p} (\mathbb{R}^d)$ of the Carnot-Caratheodory groups. However, it is not an isomorphism. One can show that its kernel consists exactly of the tree-like paths (see The Signature of a Rough Path: Uniqueness for the (involved) proof). Before continuing, let me mention that that the projective limit $G$ is an infinite-dimensional Lie group and can be identified with the character group of a Hopf algebra (see Character groups of Hopf algebras as infinite-dimensional Lie groups for an article developing the Lie group structure)
So the quotient of the group of paths (the so called reduced path group) is isomorphic to a subgroup of an infinite-dimensional Lie group. Does this imply that the rough paths of a given regularity (modulo tree like equivalences) are a Lie group itself? Unfortunately, this does not seem to be the case: On one hand the projective limit Lie group the subgroup is sitting in has very strong Lie theoretic properties which seem not to be shared by the reduced path subgroups. To illustrate this, let me mention that for a given regularity not every element in a reduced path subgroup $R$ admits a square root, i.e. in general for $g \in R$ one can not find an element $h \in R$ (the root) such that $g = h\cdot h$. (This seems to be a folklore fact in the community.) As a consequence if $R$ is a Lie group, then $g$ can not be contained in the image of its Lie group exponential (if it were, say $g = \exp_R (X)$, then $\exp_R(\frac{X}{2})$ would be a root). Now if $R$ was a (closed) Lie subgroup of $G$, then the Lie group exponentials of $R$ and $G$ would be related (this is explained nicely in Neeb's Towards a Lie theory of locally convex groups in the chapter on locally exponential Lie groups) via the inclusion map $R \rightarrow G$. However, this would imply that the Lie group exponential of $R$ needs to be a bijection as the Lie group exponential $\exp_G$ of the character group is a bijection (see Theorem B of 2). Since this contradicts the existence of group elements without roots we conclude that $R$ can not be a closed Lie subgroup of $G$. So the upshot is that the canonical Lie group structure of $G$ will not be inherited by the reduced path subgroups (though this is not saying that the reduced path subgroups could not be a Lie group in some sense, possibly with a finer topology).
This is all a beautiful theory and yields an elegant way of combining algebraic and analytic properties of rough paths in geometric language. However, as the original question is concerned: There is no known canonical Lie group structure on the space of rough paths and to my knowledge there is a priori no distinguished Lie group action (of the finite or infinite-dimensional groups the paths take their values in) which gives a deeper meaning to this.
To make it perfectly clear: Though there is no group action satisfying your curiosity, I do not want to imply by any means that it is useless to consider these paths as taking values in the groups. Indeed it is an elegant way to formulate many properties and to use (Carnot-Caratheodory) group techniques in some of the proofs in rough path theory. For example, the $1/p$-Hölder regularity condition required in rough path theory makes sense as an actual Hölder-continuity estimate with respect to the the Carnot-Caratheordory metric on the group.
