Let $G$ be a finite $p$-group, $x\in G$ and let the cyclic subgroup $\langle x \rangle$ is not normal in $G$. Let $z\in G$, $|z|=p$ and $\{\langle xz^i\rangle~|~1\leq i\leq p-1\}\cap [\langle x\rangle]_G\neq \emptyset$ . Is it true that $\{\langle xz^i\rangle~|~1\leq i\leq p-1\}\subseteq [\langle x\rangle]_G$? (Note that $p$ is prime, $|z|$ denotes the order of the element $z$ and $[\langle x\rangle]_G$ denotes the conjugacy class of the cyclic subgroup $\langle x\rangle$ in $G$)
Thank you very much for your contribution!
The answer is no.
Consider the symmetry group of the square, $D_8$. Let $r$ be a quarter rotation, and $f$ be a reflection. In keeping with your notation, let $x = r^2 f$ and $ z =f$. We have that $\langle x \rangle = \{e ,r^2 f \}$ and $\langle z \rangle = \{e ,f \}$. Moreover $\{\langle xz^i\rangle \mid 1\leq i < 2-1\} = \langle xz \rangle = \langle r^2 \rangle$.
Since $r^2$ is in the centre of $D_8$, $r^2$ is conjugate to itslef only. In particular, $r^2$ is not conjugate to $x= r^2 f$.