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2 change notation $f$ to $b$ as $f$ denotes the closed immersion $Z\to X'$.

If $Z$ is reduced, then the answer is yes, and the proof is rather elementary.

First, we can suppose $X$ is affine. Indeed, cover $X$ by affine open subsets $X_i$'s. Then $Z\cap X'_i$ is also stable by $F'_X$ because both $X'_i$ and $Z$ are stable by $F'_X$. If we can show that $Z\cap X'_i$ is defined over $k$, then $Z$ is clearly defined over $k$ too.

So suppose $X=\mathrm{Spec} A$ is affine. The closed subscheme $Z$ is defined over some finite extension $K/k$ of degree $r$. Let $I$ be the defining ideal of $Z$ in $X_K$. We have to show that $I$ is stable by $\mathrm{Gal}(K/k)$. Consider the generator $\sigma: \lambda \mapsto \lambda^{q^{r-1}}$ of $\mathrm{Gal}(K/k)$. Let $f=\sum_i b=\sum_i a_i\otimes \alpha_i\in I$ with $a_i\in A$ and $\alpha_i\in K$. Denote by $F_{X_K}$ (F_{X})_K$the relative Frobenius$F_X$of$X$extended to$K$. Then $$\sigma(f)^q=(\sum_{i} \sigma(b)^q=(\sum_{i} a_i\otimes \alpha_{i}^{q^{r-1}})^q=\sum_{i} a_i^q\otimes \alpha_{i}=(F_{X})_K(f)\in alpha_{i}=(F_{X})_K(b)\in I$$ Therefore$\sigma(f)\in \sigma(b)\in \sqrt{I}=I$and we are done. 1 If$Z$is reduced, then the answer is yes, and the proof is rather elementary. First, we can suppose$X$is affine. Indeed, cover$X$by affine open subsets$X_i$'s. Then$Z\cap X'_i$is also stable by$F'_X$because both$X'_i$and$Z$are stable by$F'_X$. If we can show that$Z\cap X'_i$is defined over$k$, then$Z$is clearly defined over$k$too. So suppose$X=\mathrm{Spec} A$is affine. The closed subscheme$Z$is defined over some finite extension$K/k$of degree$r$. Let$I$be the defining ideal of$Z$in$X_K$. We have to show that$I$is stable by$\mathrm{Gal}(K/k)$. Consider the generator$\sigma: \lambda \mapsto \lambda^{q^{r-1}}$of$\mathrm{Gal}(K/k)$. Let$f=\sum_i a_i\otimes \alpha_i\in I$with$a_i\in A$and$\alpha_i\in K$. Denote by$F_{X_K}$the relative Frobenius$F_X$of$X$extended to$K$. Then $$\sigma(f)^q=(\sum_{i} a_i\otimes \alpha_{i}^{q^{r-1}})^q=\sum_{i} a_i^q\otimes \alpha_{i}=(F_{X})_K(f)\in I$$ Therefore$\sigma(f)\in \sqrt{I}=I\$ and we are done.