Let $\mathbb{F}_q$ be an additive group of finite field and $\mathbb{F}_q' \simeq \mathbb{Z}_{q-1}$ be a multiplicative group of finite field.
I want to build a semidirect product $\mathbb{F}_q \rtimes \mathbb{F}_q'$. Every semidirect product is determined by the choise of homomoprhism $\mathbb{F}_q' \rightarrow Aut(\mathbb{F}_q)$. Such homomorphism is determined by its value on a generator of $\mathbb{F}_q'$.
We know that $\mathbb{F}_q \simeq \mathbb{Z}_{p}^d$, where $q = p^d$. Note that $Aut(\mathbb{Z}_{p}^d) \simeq GL_d(\mathbb{Z}_{p})$. Every image of a generator of $\mathbb{F}_q'$ has order, dividing $q-1$.
Thus a homomorphism $\mathbb{F}_q' \rightarrow Aut(\mathbb{F}_q)$ is determined by a choice of a matrix $A \in GL_d(\mathbb{Z}_{p})$ such that $A^{q-1} = E$. But I don't know how to get an explicit function law of such homomorphism.
This is, in a special case, the finite affine group ${\rm Aff}(q)$, which has "function law" matrix multiplication, see here for the general case. So we have the group $$ {\rm Aff}(q)=\{\begin{pmatrix} x & y \cr 0 & 1 \end{pmatrix}\mid x\in \mathbb{F}_q^{\times}, y\in \mathbb{F}_q\}, $$ of order $q(q-1)$, where the homomorphism is given by $\gamma\colon \mathbb{F}_q^{\times} \rightarrow Aut(\mathbb{F}_q)$, $\gamma_s(v)=sv$.
Edit: In general there will be other possibilities, but we still can represent the semidirect product as affine transformations.