This is well known (see e.g.  [KS06, p.381]) but we supply a proof for the convenience of the reader. To simplify notation, we set $r:=r_{{\mathcal{X}}{\mathcal{X}}^{\prime}}$, $A:=A_{{\mathcal{X}}{\mathcal{X}}^{\prime}}$, $\Delta:=\Delta({\mathcal{X}})$ and $\Delta^{\prime}:=\Delta({\mathcal{X}}^{\prime})$.

Let $W$ be the center of the blowup $\rho$, and $Z$ the smallest stratum of ${\mathcal{X}}_{0}$ containing $W$. Let $E_{i}$, $i\in I$ be the irreducible components of ${\mathcal{X}}_{0}$, $J\subset I$ the subset such that $Z$ is an component of $E_{J}$, and $\sigma_{Z}$ the simplex defined by $Z$. Let $E^{\prime}_{i}$, $i\in I$ be the strict transform of $E_{i}$ to ${\mathcal{X}}^{\prime}$. Finally, let $E^{\prime}$ be the exceptional divisor of $\rho$. It corresponds to a vertex $v^{\prime}=v^{\prime}_{E^{\prime}}$ of $\Delta^{\prime}$.

First assume $W\subsetneq Z$. In this case, $\Delta^{\prime}$ is obtained from $\Delta$ by “raising a tent over the simplex $\sigma_{Z}$”. Let us be more precise. Consider a simplex $\sigma$ of $\Delta$, corresponding to a stratum $Y$ of ${\mathcal{X}}_{0}$. By the definition of a simple blowup, $W$ meets every irreducible component of ${\mathcal{X}}_{0}$ transversely (if at all). It follows that $Y$ cannot be contained in $W$, so $\rho$ is a biholomorphism above a general point of $Y$. Thus the strict transform $Y^{\prime}$ of $Y$ defines a stratum of ${\mathcal{X}}^{\prime}_{0}$ as well as a simplex $\sigma^{\prime}$ of $\Delta^{\prime}$, whose vertices correspond to the strict transforms of the vertices of $\sigma$. In this case, $r$ maps $\sigma^{\prime}$ onto $\sigma$, and $\rho\colon Y^{\prime}\to Y$ is a bimeromorphic morphism, so $\sigma^{\prime}$ is active for $r$.

This proves that $r\colon A\to\Delta$ is surjective. To prove injectivity, consider a stratum $Y^{\prime}$ of ${\mathcal{X}}^{\prime}_{0}$, with corresponding simplex $\sigma^{\prime}$ of $\Delta^{\prime}$. If $Y^{\prime}$ is not contained in $E^{\prime}$, then $\rho$ is a biholomorphism at the general point of $Y^{\prime}$, $Y:=\rho(Y^{\prime})$ is a stratum of ${\mathcal{X}}_{0}$ of the same dimension as $Y^{\prime}$, and $Y^{\prime}$ is the strict transform of $Y$. Thus we are in the situation above. On the other hand, if $Y^{\prime}$ is contained in $E^{\prime}$, then there exist irreducible components $E_{i}$, $i\in J$ of ${\mathcal{X}}_{0}$, having strict transforms $E^{\prime}_{i}$, $i\in J$, such that $\sigma^{\prime}$ has $v^{\prime}$ and $v^{\prime}_{i}$, $i\in J$ as vertices. Since $W$ is not a stratum of ${\mathcal{X}}_{0}$, the smallest stratum $Y$ containing $\rho(Y^{\prime})$ is cut out by $E_{i}$, $i\in J$. It follows that $r$ maps the simplex $\sigma^{\prime}$ onto the lower-dimensional simplex $\sigma$, so $\sigma^{\prime}$ is not active for $r$. Hence $r\colon A\to\Delta$ is injective.

Now assume $W=Z$ is stratum of ${\mathcal{X}}_{0}$, defining a simplex $\sigma$ with vertices $v_{i}$, $i\in J$. In this case, $\Delta^{\prime}$ is obtained from $\Delta$ by a barycentric subdivision of the simplex $\sigma_{Z}$. Again, let us be more precise. The same argument as above shows that if $Y$ is a stratum of ${\mathcal{X}}_{0}$ that is not contained in $W$, and $Y^{\prime}$ is the strict transform, then the simplex $\sigma^{\prime}_{Y^{\prime}}$ is active for $r$ and $r(\sigma^{\prime}_{Y^{\prime}})=\sigma_{Y}$. Further, $\sigma^{\prime}_{Y^{\prime}}$ is the unique simplex in ${\mathcal{X}}^{\prime}_{0}$ that is active for $r$ and whose image under $r$ meets the interior of $\sigma_{Y}$.

It remains to consider strata of ${\mathcal{X}}_{0}$ contained in $Z$. This becomes a toroidal calculation. Let $Y$ be such a stratum, cut out by $E_{i}$, $i\in K$, where $J\subset K$. Then $\rho^{-1}(Y)$ consists of $|J|$ strata $Y^{\prime}_{i}$, $i\in J$, each cut out by $E^{\prime}$ and $E^{\prime}_{j}$, $j\in K\setminus\{i\}$. The restriction $\rho|_{Y^{\prime}_{i}}\colon Y^{\prime}_{i}\to Y$ is a bimeromorphic morphism, and the the corresponding simplex $\sigma^{\prime}_{i}$ is active for $r$ and maps homeomorphically onto a simplex contained in $\sigma_{Y}$. Further, these simplices $r(\sigma^{\prime}_{i})$ have disjoint interiors and cover $\sigma_{Y}$. Finally, if $Y^{\prime}$ is a stratum of ${\mathcal{X}}^{\prime}_{0}$ contained in $E=\rho^{-1}(Z)$, then $Y=\rho(Y^{\prime})$ is a stratum contained in $Z$, hence $Y^{\prime}=Y^{\prime}_{i}$ is one of the strata above. This completes the proof. ∎