Embedding $\mathrm{PGL}(n,q^h)$ in $\mathrm{PGL}(nh,q)$ It is not very hard to see that for each prime power $q$ and natural numbers $n,h$, we have an embedding
$$\iota \colon \mathrm{GL}(n,q^h) \hookrightarrow \mathrm{GL}(nh, q),$$
obtained by choosing a basis for the finite field $\mathbb{F}_{q^h}$ over the base field $\mathbb{F}_q$.
It is not very hard either to see that this does not induce an embedding of the corresponding projective groups $\mathrm{PGL}(n,q^h)$ into $\mathrm{PGL}(nh, q)$.
Of course, this does not exclude the possibility that $\mathrm{PGL}(n,q^h)$ is nevertheless isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$, and that is precisely my question:

For which values of $q,n,h$ with $h > 1$ is $\mathrm{PGL}(n,q^h)$ isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$?

Notice that comparing the order of both groups does not rule out anything at all.
I believe I read somewhere that $\mathrm{PGL}(n,q^h)$ is isomorphic to a subgroup of $\mathrm{PGL}(n^h, q)$ (but I forgot the precise construction), so this would at least tell that the answer is positive for $n=h=2$. (Edit: This fact is now confirmed by Derek Holt's answer.)
 A: Yes, ${\rm PGL}(2,q^2)$ is a subgroup of ${\rm PGL}(4,q)$, but I would guess that that is an exception, and in general there is no such embedding.
${\rm GL}(4,q)$ contains the subgroup that I denote by ${\rm CO}^-(4,q)$, which is the conformal orthogonal group of minus-type (and equal to the normalizer in ${\rm GL}(4,q)$ of ${\rm GO}^-(4,q)$).
The projective image ${\rm PCO}^-(4,q)$, which is of course a subgroup of ${\rm PGL}(4,q)$, happens to be isomorphic to ${\rm P \Gamma L}(2,q^2)$, which contains ${\rm PGL}(2,q^2)$ as a subgroup of index $2$.
So this is a separate embedding, and is not related to the semiliear embedding ${\rm GL}(2,q^2) \to {\rm GL}(4,q)$.
The general embedding ${\rm PGL}(n,q^h) \to {\rm PGL}(n^h,q)$ can be defined as follows. Let $M$ be the natural module for $G={\rm GL}(n,q^h)$ and let $\phi$ be the field automorphism of $G$ induced by the automorphism $x \mapsto x^q$ of the field. Then $M \otimes M^\phi \otimes M^{\phi^2} \otimes \cdots \otimes M^{\phi^{h-1}}$ is a module of dimension $n^h$ that is stabilized by $\phi$ and hence can be realized over ${\mathbb F}_q$. So we get a homomorphism ${\rm GL}(n,q^h) \to {\rm GL}(n^h,q)$, which is not always injective, but scalars map to scalars, so it induces the required embedding ${\rm PGL}(n,q^h) \to {\rm PGL}(n^h,q)$.
A: Assume that $n$ is prime to $q^h-1$. Then $\mathrm{SL}(n, q^h)\cap \mathbb F_{q^h}^{\times}$ is trivial and the map $\mathrm{SL}(n, q^h)\rightarrow \mathrm {PGL}(n, q^h)$ is an isomorphism, as $x\mapsto x^n$ induces an iso on $\mathbb{F}_{q^h}^{\times}$. The composition 
$$ \mathrm{SL}(n, q^h)\rightarrow \mathrm{GL}(n, q^h)\rightarrow \mathrm{GL}(nh, q)\rightarrow \mathrm{PGL}(nh, q),$$
where the middle arrow is the one described in the question, is injective, because
the kernel is equal to $\mathrm{SL}(n, q^h)\cap \mathbb F^{\times}_{q}$.
