If we let $k(G)$ denote the number of conjugacy classes of the finite group $G$,
then for any subgroup $H$ of $G$ (normal or not), a Theorem of P.X. Gallagher
states that $[G:H]^{-1}k(H) \leq k(G) \leq [G:H]k(H)$ (this has probably been discovered
and rediscovered many times). I find that the easiest way to see it is using irreducible
complex characters. Of course, $k(G)$ is also the number of complex irreducible characters of $G$,
and likewise for $H$. For each irreducible character $\chi$ of $G$, there is an irreducible
character $\mu$ of $H$ such that $\mu$ occurs with non-zero multiplicity in the restriction of $\chi$ to $H$. By Frobenius reciprocity, $\chi$ is an irreducible constituent of the character of $G$
induced from the character $\mu$ of $H$. On the other hand, using Frobenius reciprocity again, each irreducible constituent of $\mu$ induced to $G$ must have degree at least $\mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of $\mu$ induced to $G$
Since this is true for each irreducible character of $H$, and since each irreducible character of $G$ must appear with non-zero multiplicity in at least one such character, we have
$k(G) \leq [G:H]k(H).$ Going in the other direction, if $\chi$ is an irreducible character of $G$
and $\mu$ is an irreducible constituent of te restriction of $\chi$ to $H$, then we have
$\chi(1) \leq [G:H]\mu(1)$, since $\chi$ occurs as a constituent of $\mu$ induced to $G$.
Thus $\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible
constituents of the restriction of $\chi$ to $H$. Since each irreducible character of $H$ occurs
as a constituent of some such irreducible character of $G$ (consider, for example, the restriction
of the regular character), we have $k(H) \leq [G:H] k(G).$