= Solution
Because $N$ is an <abelian group>, every $\varphi\in\operatorname{Irr}(N)$ is a <linear character>, hence a homomorphism to $\mathbb C^\times$. Its <inertia group of a character> contains $N$, whose inner conjugations act trivially on $N$. Write an arbitrary $g\in G$ as $g=nh$ with $n\in N$ and $h\in H$, using $G=NH$. Conjugation by $n$ does not affect $\varphi$, so $g$ stabilizes $\varphi$ if and only if $h$ does.
Suppose $h\ne1$ stabilizes $\varphi$. Then $\varphi(h^{-1}yh)=\varphi(y)$ for every $y\in N$, and its multiplicativity gives
$$
\varphi([h,y])=\varphi(h^{-1}y^{-1}hy)=\varphi(h^{-1}yh)^{-1}\varphi(y)=1.
$$
Part (iv) makes $y\mapsto[h,y]$ surjective onto $N$, so $\varphi(x)=1$ for every $x\in N$. This would be the <trivial character>, contrary to the hypothesis. Therefore no nonidentity $h\in H$ stabilizes $\varphi$, and
$$
\boxed{I_G(\varphi)=N\qquad(\varphi\ne1_N).}
$$
The argument uses both abelianness, to make $\varphi$ multiplicative, and the fixed-point-free commutator bijection.
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