Problem- Let $q$ be a power of a prime $p$ say $q=p^k$. Show that there exists a group $G$ of order $q(q-1)$ with a normal elementary abelian subgroup of order $q$ and such that all elements of order $p$ in $G$ are conjugate.
I tried it, but failed. let me write here what i thought
$\textbf{Try 1}$ - If i consider $G=(C_p$$\times C_p\times C_p...\times C_p)$$\times C_{q-1}$ where $C_p$'s direct product is $k$ times. Now it is a group of order $q(q-1)$ and also has a normal elementary abelian subgroup of order $q$ but doesn't satisfy all elements of order $p$ in $G$ are conjugate.
May be if I use semidirect products as $G=(C_p$$\times C_p\times C_p...\times C_p)\rtimes_\phi M$ for some group $M$ possibly non abelian of order $q-1$, Is there hope with this, this will need at least an hour of thinking and trying , and choosing a good $\phi$ too?
$\textbf{Try 2}$ - As we know that a minimal normal subgroup of finite solvable group is elementary abelian, so if i can look for a solvable group of order $q(q-1)$ s.t. it has a minimal normal subgroup of order $q$ i am half done, now "elements of order $p$ should be conjugate" is causing trouble. It may be satisfied here, but i am not sure.
Any help is appreciated.
Your Try 1 works, but you need to use a semidirect product.
To see how/why, let $C_p\times C_p\times\cdots \times C_p$ be the additive group of the finite field $\Bbb{F}_q$. Also let $C_{q-1}$ be the multiplicative group of the same field (known to be cyclic of order $q-1$). If we let $\Bbb{F}_q^*$ act on $\Bbb{F}_q$ by the field multiplication, then all the non-zero elements of $\Bbb{F}_q$ are in the same orbit of that action. The semidirect product then turns that orbit into a conjugacy class.
If you prefer another model (isomorphic to the previous one, but possibly easier to handle), try the group of matrices $$ G=\left\{\left(\begin{array}{cc}\alpha&\beta\\0&1\end{array}\right)\,\bigg\vert\,\alpha,\beta\in\Bbb{F}_q,\alpha\neq0\right\}. $$ You may also know this as the group of affine linear transformations of the line $\Bbb{F}_q$.