The $G$ action on $G^{\ast n}$ seems to be somewhat complicated: here's a description, but I hope someone can find a simpler one.

Let $G$ be a finite group. $G^{\ast n}$ can be described as a simplicial complex with vertex set $G \times [n]$ (where $[n] = \{1, 2, \ldots, n\}$), whose maximal simplices are sets of the form $\{(a_1, 1), (a_2, 2), \ldots, (a_n, n)\}$ for some $(a_1, \ldots, a_n) \in G^n$ ("pick one element of $G$ from each column of $G \times [n]$").

The union $T$ of the maximal simplices where at least one $a_i = e$ (where $e$ is the identity of $G$) is a contractible space --it can easily be retracted onto the simplex $\{(e,1), (e,2), \ldots, (e,n)\}$. The quotient $G^{\ast n}/T$ is then a wedge of copies of $S^{n-1}$, one for each of the $(|G|-1)^n$ maximal simplices that don't have any $a_i=e$. The $S^{n-1}$ corresponding to the maximal simplex $\{(a_1, 1), \ldots, (a_n, n)\}$ is $\{e, a_1\} \ast \{e, a_2\} \ast \cdots \ast \{e, a_n\} \subset G^{\ast n}$.

(I had first computed the homotopy type as Jesús did in this comment, but that didn't seem to make it easy to figure out the action.)

On $G^{\ast n}$ the $G$-action is straightforward, but let's see what it corresponds to in the quotient $G^{\ast n}/T$. For each $\mathbf{a} = (a_1, \ldots, a_n)$, let $\sigma(\mathbf{a})$ be the corresponding $(n-1)$-sphere in $G^{\ast n}/T$. The preimage of $\sigma(\mathbf{A})$ in $G^{\ast n}$ has maximal simplices $\{(a'_1, 1), \ldots, (a'_n,n)\}$ where each $a'_i$ is either $a_i$ or $1$. Applying the action of $g \in G$, this preimage goes to the $(n-1)$-sphere with faces $\{(ga'_1, 1), \ldots, (ga'_n,n)\}$. In $G^{\ast n}$ the faces with some $ga'_i=e$ get contracted to the basepoint, and the others will wrap around $\sigma(g \cdot \mathbf{a}')$ with degree $\pm 1$.

Some care is needed to figure out the signs, but I think they work out as follows: $$g|_{\sigma(\mathbf{a})} \simeq \sum_{I \subseteq [n]} (-1)^{|I|} \sigma\left(g \cdot (I \ominus \mathbf{a})\right),$$
where:

- $I \ominus \mathbf{a} \in G^n$ is $\mathbf{a}$ with all $a_i$ with $i \in I$ replaced by $e$,
- $g \cdot \mathbf{a}'=(ga'_1, \ldots ga'_n)$,
- the sum is to be interpreted as follows: the domain
$\sigma(\mathbf{a})$ is first mapped to $\bigvee_{I \subset [n]}
S^{n-1}$ by a (multiple-)pinch map, and then the $I$-th "wedgeand" maps to
$\sigma(g \cdot I \ominus \mathbf{a})$ with degree $(-1)^{|I|}$,
except that if any coordinate of $g \cdot I \ominus \mathbf{a}$ is
$e$, the entire $I$-th wedgeand is sent to the basepoint of
$G^{\ast n}/T$.

I'm not sure you can really strictify this to an honest $G$-action on $G^{\ast n}/T$, but at the very least, the above describes for each $g \in G$ an endomap on $G^{\ast n}/T$ such that the square $$\require{AMScd}\begin{CD} G^{\ast n} @>{g}>> G^{\ast n} \\ @V{\pi}VV @V{\pi}VV \\ G^{\ast n}/T @>{g}>> G^{\ast n}/T \end{CD}$$ commutes up to homotopy.

Notice that in the case $G$ is cyclic of order $2$, with generator $\tau$, there is only one $(n-1)$-sphere, namely $\sigma(\tau, \ldots, \tau)$ and in the expression for the action of $\tau$ only the term with $I = [n]$ is not crushed to the basepoint, which says that $\tau$ acts on $\sigma(\tau, \ldots, \tau)$ with degree $(-1)^n$ --which agrees, as expected, with the antipodal action.

**ADDED LATER**: Ben Williams figured out that the somewhat messy formula above with $I \ominus \mathbf{a}$ in it meant that $G$ acts on $H_{n-1}(G^{\ast n})$ by the $n$-th tensor power of the reduced regular representation and asked me if there was a space level explanation of that. Here's my explanation: if one is willing to settle for the homotopy type of the suspension of $G^{\ast n}$, we have that $G$-equivariantly $\Sigma G^{\ast n} \cong (SG)^{\wedge n}$, where $SG$ denotes the unreduced suspension of $G$; and $SG \cong \bigvee_{|G|-1} S^1$ with $H_1(SG)$ being the reduced regular representation.