Normal groups are necessarily solvable. Indeed, by Jordan-Holder decomposition an unsolvable finite group $G$ necessarily contains a normal subgroup $H$ that maps surjectively to a nonabelian finite simple group $S$. Take $H$ minimal with this property.

$S$ is not a $p$-group for any $p$, hence its order has two distinct prime factors $p_1$ and $p_2$. By Cauchy's theorem, it contains elements $g_1$ and $g_2$ of orders $p_1$ and $p_2$. Let $r_1$ be an inverse image of $g_1$ in $H$, chosen to have $p_1$-power order (possible by taking an arbitrary list and taking a suitable prime-to-$p_1$ power of it). Let $r_2$ be the same for $g_2$.

Then the normal subgroup generated by all the conjugates of $r_1$ is $H$. Indeed, it is manifestly contained in $H$, as $H$ is normal and contains $r_1$. Its image under the projection to $S$ is some normal subgroup of $S$ containing $g_1$, hence all of $S$. By minimality of $H$, it is $H$. Then the same is true for $r_2$.

But $r_1$ and $r_2$ have distinct orders, hence the cyclic subgroups they generate cannot be conjugate.

However, normal groups need not be nilpotent. Consider the group of all affine transformations of $\mathbb F_q$, aka $\mathbb F_q \rtimes \mathbb F_q^{\times}$. All elements of order $p$ are conjugate and generate the subgroup $\mathbb F_q$, and all nontrivial elements of prime-to-$p$ order $n$ are conjugate and generate the inverse image of the subgroup of $n$th roots of unity in $\mathbb F_q^{\times}$.

Normal groups are necessarily solvable. Indeed, by Jordan-Holder decomposition an unsolvable finite group $G$ necessarily contains a normal subgroup $H$ that maps surjectively to a nonabelian finite simple group $S$. Take $H$ minimal with this property.

$S$ is not a $p$-group for any $p$, hence its order has two distinct prime factors $p_1$ and $p_2$. By Cauchy's theorem, it contains elements $g_1$ and $g_2$ of orders $p_1$ and $p_2$. Let $r_1$ be an inverse image of $g_1$ in $H$, chosen to have $p_1$-power order (possible by taking an arbitrary lift and taking a suitable prime-to-$p_1$ power of it). Let $r_2$ be the same for $g_2$.

Then the normal subgroup generated by all the conjugates of $r_1$ is $H$. Indeed, it is manifestly contained in $H$, as $H$ is normal and contains $r_1$. Its image under the projection to $S$ is some normal subgroup of $S$ containing $g_1$, hence all of $S$. By minimality of $H$, it is $H$. Then the same is true for $r_2$.

But $r_1$ and $r_2$ have distinct orders, hence the cyclic subgroups they generate cannot be conjugate.

However, normal groups need not be nilpotent. Consider the group of all affine transformations of $\mathbb F_q$, aka $\mathbb F_q \rtimes \mathbb F_q^{\times}$. All elements of order $p$ are conjugate and generate the subgroup $\mathbb F_q$, and all nontrivial elements of prime-to-$p$ order $n$ are conjugate and generate the inverse image of the subgroup of $n$th roots of unity in $\mathbb F_q^{\times}$.