There is the following adjointness (a form of Frobenius reciprocity):
$Hom_G(rho,F^X) = Hom_H(rho,trivial).$
Thus $rho$ embeds in $F^X$ if and only if $rho$ admits a non-trivial $H$-fixed quotient.
(If $H$ is finite and $F$ has characteristic zero, or at least prime to the order
of $H$, so that $rho$ is semi-simple
as an $H$-representation, then this is equivalent to requiring that $rho$ have a
non-trivial $H$-fixed subrepresentation.)
(Note also that a non-zero $G$-equivariant map out of $rho$ is automatically injective,
because $rho$ is irreducible.)
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