We define the holomorph of a group, $\operatorname{Hol}(G)$, as its semidirect product $G\rtimes _f\operatorname{Aut}(G)$. As it happens (as is shown here), $D_4\approx\operatorname{Aut}(D_4)$, and we can say that much like how $D_4 = \langle r,s\ |\ r^4=s^2=e,\ srs=r^3\rangle$, we have mappings $\sigma: r\mapsto r^3$ and $\rho^i: s\mapsto sr^i$ such that $\operatorname{Aut}(D_4)=\langle\rho,\sigma\ |\ \rho^4=\sigma^2=e,\ \sigma\rho\sigma=\rho^3 \rangle$.
With this in mind, we can now discuss $\operatorname{Hol}(D_4)$. As it is defined as $D_4\rtimes_f\operatorname{Aut}(D_4)4$, we have a mapping $f: D_4\rightarrow\operatorname{Aut}(D_4)\approx\operatorname{Aut}(D_4)\rightarrow D_4$ defined as $f_h(n)=hnh^{-1}$, with $h\in D_4$ and $n\in\operatorname{Aut}(D_4)$. One mapping of note here is $f_s:\rho\mapsto\rho^3$.
Is this isomorphic to any familiar group? As it is the semidirect product of two groups, by definition, $|\operatorname{Hol}(D_4)|=|D_4||\operatorname{Aut}(D_4)|=8\times8=64$. This is clearly a non-abelian group, so that restricts our field of search from 267 groups to "only" 256. Is this isomorphic to a less obscure group, or is this as good as it gets?