I will restrict to the case that $G$ is finite. There are many interesting results of this type already in the literature, some well-known, some not so well-known (some repeatedly re-discovered, so I will do my best with attributions). Of great relevance are Clifford's theorem, Frobenius reciprocity, and Brauer's permutation lemma, among other things.

Brauer's permutation lemma plays a role in changing this to a question about complex irreducible characters. Let $N$ be a normal subgroup of a finite group $G$. Then $G$ acts by conjugation on $N$, and there is a natural action of $G$ on (complex) irreducible characters of $N$ (in both cases, $N$ is acting trivially, and the action is really one of $G/N$). Brauer's permutation lemma tells us that for any $x \in G$, the number of conjugacy classes of $N$ fixed by $x$ is the same as the number of irreducible characters fixed by $x$. The usual orbit counting formula then tells us that $G$ has the same number of orbits on conjugacy classes of $N$ as it does on irreducible characters of $N$.

Hence the number of conjugacy classes of $G$ contained in $N$ is the number of $G$ orbits on irreducible characters of $N$.

Before I continue, I will fix some notation. Let $G$ be a finite group, $H$ be a (not necessarily normal) subgroup of $G$ and $N$ be a normal subgroup of $G$. We let $k(G)$ denote the number of conjugacy classes of $G$, and we let $k_{G}(N)$ denote the number of $G$-orbits on conjugacy classes of $N$, which is the same as the number of conjugacy classes of $G$ contained in $N$. Notice that it is not generally true that
$k(N) = k_{G}(N).$

Frobenius reciprocity is relevant because (even in the case that $H$ is not normal in $G$ ), we may note that for each irreducible character $\mu$ of $H$, ${\rm Ind}_{H}^{G}(\mu)$ has at most $[G:H]$ irreducible constituents (counting multiplicity), while each irreducible character $\chi$ of $G$ occurs as an irreducible constituent of some such induced character. It follows that $k(G) \leq [G:H]k(H)$.

Similarly, we find that if $\chi$ is an irreducible character of $G$, then ${\rm Res}^{G}_{H}(\chi)$ has at most $[G:H]$ irreducible constituents, so we obtain
$k(H) \leq [G:H]k(G).$

Hence we have $\frac{k(H)}{[G:H]} \leq k(G) \leq [G:H]k(H)$ whenever $H$ is a subgroup of the finite group $G$. These inequalities were proved by P.X. Gallagher around 1962. We will see that they are also relevant to the application of Clifford's theorem, as Gallagher himself knew.

Clifford's theorem is relevant for the following reason. If $\chi$ is an irreducible character of $G$, and $N \lhd G$, then the irreducible constituents of ${\rm Res}^{G}_{N}(\chi)$ lies in a single $N$-orbit (and all occur with equal multiplicity). Furthermore,
if $\mu$ is an irreducible character of $N$, then there is a bijection between irreducible characters $\chi$ of $G$ such that $\mu$ occurs with non-zero multiplicity in ${\rm Res}^{G}_{N}(\chi)$, and irreducible characters of the $G$-stabilizer of $\mu$, usually denoted $I_{G}(\mu)$, which restrict to $N$ as a multiple of $\mu$.

Using projective representations (in Schur's sense), it can be proved that the number of irreducible characters of $I_{G}(\mu)$ which restrict to a multiple of $\mu$ is at most $k(I_{G}(\mu)/N).$ The main idea is to reduce to the case that $N$ is central and $\mu$ is a linear character of $N$ by another part of Clifford's theorem(s).

Applying Gallagher's inequalities with $I_{G}(\mu)/N$ in the role of $H$, and $G/N$ in the role of $G$, , we obtain that
$k(I_{G}(\mu)/N) \leq [G:I_{G}(\mu)]k(G/N).$ Notice that $[G:I_{G}(\mu)]$ is the length of the $G$-orbit of $\mu$. Letting $\mu$ run through orbit representatives for the action of $G$ on irreducible characters of $N$, we obtain another inequality of P.X. Gallagher (also proved independently by H. Nagao around 1962): $k(G) \leq k(N)k(G/N).$

A variant of the above argument which is sometimes useful, and is relevant to the present question was proved by L.G. K'ovacs and myself in 1993: if $N$ is a normal subgroup of $G$, then we have $k(G) \leq m k_{G}(N)$, there $m$ is the maximum number of conjugacy classes of any subgroup of $G/N$. Notice that if $G/N$ is Abelian,
then $m = [G:N] = k(G/N).$ In particular, when $G/N$ is Abelian, we obtain a sharpening of the Gallagher-Nagao bound, for then we have $k(G) \leq k(G/N)k_{G}(N).$

Recalling that $k_{G}(N)$ is the number of $G$-conjugacy classes contained in $N$, we see that whenever $N$ is a normal subgroup of $G$ with $G/N$ Abelian, then we have
$k(G) \leq k(G/N)$ $\times$ (the number of conjugacy classes of $G$ contained in $N$).
This answers the question in the case $[G:N] =2$

Notice that when $N$ is a normal subgroup of $G$ with $G/N$ non-Abelian, it may not be straightforward to obtain the precise value of $m$, the maximum number of conjugacy classes of any subgroup of $G/N$ (though of course we always have $m < [G:N]$ given that $G/N$ is non-Abelian. Later edit: in fact, it is an easy exercise to check that a non-Abelian group $X$ has at most $|X|-3$ conjugacy classes, and if it has $|X|-3$ conjugacy classes, then $|X| \in \{6,8 \}).$

is"well known" but I'm not sure off the top of my head the best place to look. $\endgroup$becauseit's straightforwardly equivalent to that better-known identity (attributed to Euler in Andrews's book) even though this formulation is in some ways more natural. $\endgroup$3more comments