$\GF(p^d)\subseteq\GF(p^n)$ iff $d$ divides $n$

Suppose first that \[ \FF_{p^d}\subseteq\FF_{p^n}. \] Then the extension degree satisfies \[ [\FF_{p^n}:\FF_p] =[\FF_{p^n}:\FF_{p^d}] [\FF_{p^d}:\FF_p]. \] Thus \[ n=[\FF_{p^n}:\FF_{p^d}]\,d, \] so \(d\mid n\).

Conversely, suppose \(d\mid n\), say \(n=dm\). Every element of \(\FF_{p^d}\) satisfies \[ x^{p^d}=x. \] Hence \[ x^{p^n}=x^{(p^d)^m}=x, \] so every element of \(\FF_{p^d}\) is a root of \(x^{p^n}-x\). The roots of this polynomial in an algebraic closure are exactly the elements of \(\FF_{p^n}\). Therefore \[ \FF_{p^d}\subseteq\FF_{p^n}. \]