A marked ideal is equivalent to its powers

Exact order of powers. Fix $x\in X$, put $R={\mathcal{O}}_{X,x}$ and let $\mathfrak{m}$ be its maximal ideal. If ${\mathcal{I}}_x=0$, then every positive power is zero and both orders are $+\mathrm{\infty }$. Otherwise $a={\mathrm{ord}}_x(\mathcal{I})$ is finite and attained by [F1]. Choose $f\in {\mathcal{I}}_x\setminus {\mathfrak{m}}^{a+1}$; its initial class in ${\mathrm{gr}}_{\mathfrak{m}}R$ is nonzero. By [A1], [F2] and [F3], this associated-graded ring is a domain, so the initial class of $f^k$ is nonzero in degree $ka$. Thus ${\mathrm{ord}}_x({\mathcal{I}}^k)\le ka$, while ${\mathcal{I}}_x\subseteq {\mathfrak{m}}^a$ gives the reverse inequality. Hence ${\mathrm{ord}}_x({\mathcal{I}}^k)=k{\mathrm{ord}}_x(\mathcal{I})$.

Equal supports. By step 1.1, ${\mathrm{ord}}_x(\mathcal{I})\ge \mu$ if and only if ${\mathrm{ord}}_x({\mathcal{I}}^k)=k{\mathrm{ord}}_x(\mathcal{I})\ge k\mu$. Therefore $\mathrm{supp}(\mathcal{I},E,\mu )=\mathrm{supp}({\mathcal{I}}^k,E,k\mu )$, including $\mu =0$.

Equal multiple test blow-ups. Induct on the sequence length. At each stage $i$, assume the two transforms are $({\mathcal{I}}_i,E_i,\mu )$ and $({\mathcal{I}}_i^k,E_i,k\mu )$. Step 1.1 gives equal supports for these marked ideals, so their admissible regular centers meeting $E_i$ with SNC agree. If $y$ is the exceptional equation for such a center, the controlled-transform product identity [F4] gives $y^{-k\mu }{\sigma }^{\ast }({\mathcal{I}}_i^k)=(y^{-\mu }{\sigma }^{\ast }{\mathcal{I}}_i{)}^k$, so the next transforms again have this form and their supports agree. The base case is step 2.1; induction works in both directions, so the test sequences and all induced supports coincide. By [F5], the marked ideals are equivalent.