$x^{p^d}-x$ divides $x^{p^n}-x$ iff $d$ divides $n$

Work in \(\mathbb F_p[x]\). Factor \[ x^{p^m}-x=x\bigl(x^{p^m-1}-1\bigr). \] Since the ring is a domain, cancellation of the common factor \(x\) gives \[ x^{p^d}-x\mid x^{p^n}-x \iff x^{p^d-1}-1\mid x^{p^n-1}-1. \] For positive integers \(A,B\) over any field, \[ x^A-1\mid x^B-1\iff A\mid B. \] Indeed, if \(B=qA+r\) with \(0\le r<A\), division by \(x^A-1\) leaves remainder \(x^r-1\), which vanishes exactly when \(r=0\).

Hence \[ x^{p^d}-x\mid x^{p^n}-x \iff p^d-1\mid p^n-1. \] Finally, for integers \(p>1\), \[ p^d-1\mid p^n-1\iff d\mid n, \] by the same Euclidean-division argument on the exponents. Therefore \[ \boxed{x^{p^d}-x\mid x^{p^n}-x\iff d\mid n}. \]