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Let $G$ be a group. Let $(\Pi,\circ)$ be a groupoid. Suppose I have a $G$-action on every morphism space $\Pi(p,q)$, denoted by $G\times \Pi(p,q)\to \Pi(p,q)$, $(g, \sigma)\mapsto g\cdot \sigma$. (For simplicity, we may assume $G$ is abelian.) Suppse these $G$-actions further satisfy the natural properties (1) $g\cdot (\sigma_1\circ \sigma_2)=(g\cdot \sigma_1)\circ \sigma_2=\sigma_1\circ (g\cdot \sigma_2)$; (2) $g_1g_2\cdot \sigma=g_1\cdot(g_2\cdot \sigma)$; (3) $(g\cdot \sigma)^{-1}=g^{-1}\cdot \sigma^{-1}$ (I might miss some.)

Is there a nice conceptual way to describe the above situation? I think what I ask is standard, but I fail to find a reference.

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    $\begingroup$ Isn't this a groupoid enriched in $G$-sets? $\endgroup$
    – LSpice
    Commented Jun 13, 2021 at 1:06
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    $\begingroup$ @Hang it's a cartesian monoidal category, where the product of $G$-sets is the cartesian product with the diagonal action. Also, if you are thinking of this as the group acting on the groupoid, then you can write it as $G\to \mathrm{Aut}(\Pi)$, where $\mathrm{Aut}(\Pi)$ is the subset of endofunctors of $\Pi$ that are isomorphisms (a group by composition of functors), and this homomorphism factors through the subgroup where the functors are the identity on objects. $\endgroup$
    – David Roberts ♦
    Commented Jun 13, 2021 at 1:27
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    $\begingroup$ It seems to me that the compatibility with composition asked for here is NOT the one relating to the standard monoidal structure David explains. Otherwise we would have $g\cdot(\sigma_1\circ \sigma_2) = (g\cdot \sigma_1)\circ (g\cdot \sigma_2)$ $\endgroup$
    – Achim Krause
    Commented Jun 13, 2021 at 8:44
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    $\begingroup$ Property (1) even tells you that the action factors through the abelianisation of $G$ (you can show $(gh)\cdot (\sigma\circ id) = (hg) \cdot (\sigma\circ id) $), so it seems it's closer to the monoidal structure referred to by Zhen Lin. $\endgroup$
    – Achim Krause
    Commented Jun 13, 2021 at 8:56
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    $\begingroup$ Let $BG$ denote the category on a single object with automorphisms $G$. Note that $G$ is abelian if and only if $BG$ is monoidal. In this case, $BG$ is a group object in the category of groupoids, and what the OP describes as $\Pi$ is precisely a monoid object for $BG$ in the category of groupoids. $\endgroup$
    – Konrad Waldorf
    Commented Jun 14, 2021 at 10:39

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