= Solution
{c}
Suppose an arbitrary <binary operation> $\star$ makes $\mathbb Z$ a <group> and makes $\rho$ a left <group action> on $G$. For any <group action>, each acting element induces a <bijection>: the <function> induced by its <inverse element> is its <inverse function>.
In particular, the acting element whose underlying integer is $0$ must induce a <bijection> $G\to G$. But its power <function> is
$$
x\longmapsto\rho(0,x)=x^0=e,
$$
a constant <function>. This is <injective> only when all elements of $G$ equal $e$. Therefore
$$
\boxed{G=\{e\}.}
$$
This proves the <power-map group action obstruction> for every possible $\star$; it never assumes that $\star$ is ordinary integer <multiplication>, or that the integer $0$ is the acting <identity element>.
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