This is to understand better Example 3.9 on page 221 of *Group actions on stacks and applications* by M.Romagny.

For an action of an algebraic group (scheme) $G$ on an algebraic stack $\mathcal{M}$, the author defines a fixed point stack $\mathcal{M}^G$ as follows. Over a base scheme $T$, objects of $\mathcal{M}^G(T)$ are pairs

$$(x,\{\alpha^g\}_{g\in G(T)})$$

with $x\in\mathcal{M}(T)$ and $\{\alpha^g\}$ a system of (iso)morphisms in $\mathrm{mor}\mathcal{M}$ $$\alpha^g:x\to g.x$$

such that, for every $g,g'\in G(T)$, the following "non abelian group cohomology cocycle condition" holds

$$(g.\alpha^{g'})\circ\alpha^g=\alpha^{gg'}$$

and morhisms in $\mathcal{M}^G(T)$ are the obvious ones.

Now, the (counter)example in question is to show that, in general, "taking the fixed point stack" does not commute with "taking the coarse moduli space" (when the latter exists). It works as follows.

Let $H=\{\pm1,\pm i, \pm j, \pm k \}$ be the $8$ element quaternion group and $1\to Z \to H \to G \to 1$ be the quotient by its center $Z=\{\pm 1\}\simeq \mathbb{Z}/2$. The abelian group $G\simeq \mathbb{Z}/2\times \mathbb{Z}/2$ acts by conjugation on $H$, and this induces an action $G \curvearrowright BH$ , where $BH=[*/H]$ is the classifying stack of $H$.

The point of the example is that $((BH)^{\mathrm{cms}})^G=*^G=*$, while already $(BH)^G=\textrm{Ø}$.

What is not clear to me is this latter equality $(BH)^G=\textrm{Ø}$.