Let $G$ be a finite group, let $H$ be its normal subgroup of prime index. Take $x\in H$ such that the centralizer of $x$ in group $H$: $C_H(x)$ be a a proper subgroup of the centralizer of $x$ in $G$: $C_G(x)$. Then, if $y\in H$ is conjugate to $x$ in $G$, then $y$ is conjugate to $x$ in H.
I don't know how to tackle this problem. Any hints?
Note that $|x^G|=\frac{|G|}{|C_G(x)|}$, $|x^H|=\frac{|H|}{|C_H(x)|}$, $|G|=p|H|$, $|C_H(x)|<|C_G(x)|$, where $x^G$ is a conjugacy class of $x$ in $G$.
We also know that: $G > H \geq C_H(x)$ and $G\geq C_G(x)>C_H(x)$.
Let me write $p$ for the index of $H$ in $G$ and $q$ for the index of $C_H(x)$ in $C_G(x)$.
Consider the projection homomorphism from $G$ to its quotient $G/H$, and restrict that homomorphism to $C_G(x)$. That restriction $\rho$ has kernel $C_G(x)\cap H=C_H(x)$, which by assumption is not all of $C_G(x)$. So the image of $\rho$ is a non-trivial subgroup of $G/H$. But $G/H$ has prime order, so the image of $\rho$ is all of $G/H$. That is, $\rho$ induces an isomorphism from $C_G(x)/C_H(x)$ onto $G/H$; so these two groups have the same order, i.e., $p=q$.
Now remember that the number of conjugates of an element in a group is the index of its centralizer. So if $x$ had strictly more conjugates in $G$ than in $H$, then we'd have $|G|/|C_G(x)|>|H|/|C_H(x)|$, which is equivalent to $p>q$ and thus contradicts what was proved above.