Derek Holt gave an counter example of my question because i had forgotten to write this fact that $p,q$ are distinct primes. Thus i repair my question.
Let $G=P \ltimes Q$ be a finite group scuch that $P$ is a maximal subgroup of $G$ which $P$ is non-normal cyclic and $P=C_G(P)=N_G(P)$ and $G'=Q \cong C_q \times C_q \times C_q$, where $p,q$ are distinct primes. . Also assume that $G$ has the properties that if $H,K$ be two distinct non-normal non-cyclic subgroup of $G$, then $H,K$ are conjugate in $G$.
We see that there exists a homomorphism $\varphi : P \longrightarrow Aut(Q)\cong GL(3,q)$. also clearly, $Ker(\varphi)=C_P(Q)=Z(G)$. I want to know why $P \cong C_p$, where $p=q^2+q+1$ is a prime.
Here is a counterexample with $q=11$. The group ${\rm GL}(3,11)$ has a unique conjugacy class of cyclic subgroups of order $133 = 7 \times 19$.
Let $Q$ be elementary abelian of order $11^3$, let $P \cong C_p$ with $p=133$, and let $G = P \ltimes Q$ with the action corresponding to this cyclic subgroup of ${\rm GL}(3,q)$. The action of $P$ on $Q$ is irreducible, so $P$ is maximal in $G$ and hence $P=C_G(P)=N_G(P)$. Clearly $G'=Q$.
So we have to check the condition that all non-normal non-cyclic subgroups are conjugate in $G$. Let $H$ be a non-cyclic subgroup of $G$. Since $G/Q$ is cyclic, $H \cap Q \ne 1$.
But all nontrivial subgroups of $P$ (i.e. $P$ itself and its subgroups of orders $7$ and $19$) act irreducibly on $Q$, so either $Q \le H$ or $H \le Q$, If $Q \le H$ then $H \unlhd G$, so we can assume that $H < Q$. Then $H$ non-cyclic implies that $|H|=q^2$.
Now $Q$ has exactly $q^2+q+1=133$ subgroups of order $q^2$. Since $P$ and its nontrivial subgroups act irreducibly on $Q$, they cannot normalize $H$, so these $133$ subgroups are all conjugate under the action of $P$.
Note that the fact that $Q$ has $q^2+q+1$ subgroups of order $q^2$, which must all be conjugate under the action of $P$, means that $|P|$ must be divisible by $q^2+q+1$.
If $|P| > q^2+q+1$ then by the Orbit-Stabilizer Theorem, some nontrivial element $g$ of $P$ must stabilize (i.e. normalize) some subgroup $R$ of order $q^2$. Now $\langle g,R \rangle$ is a non-normal subgroup of $G$ of order greater than $q^2$, so it cannot be conjugate to the subgroups of order $q^2$, contradicting the hypotheses. So $|P|=q^2+q+1$.