In this answer I will try to explain in details the proof of
Proposition 25 in Serre's book. For my convenience in writing this,
I will use Serre's notation so my $A$ is your $N$, i.e.
$G=H\ltimes A$ and $G_i=H_i\ltimes A$ where $H_i=\{h\in H:h\chi_i=\chi_i\}$
subgroup of $H$.
(Serre) The restriction of $\theta_{i,\rho}$ to $A$ only
involves characters $\chi$ belonging to the orbit $H\chi_i$
of $\chi_i$. This shows that $\theta_{i,\rho}$ determines $i$.
As the statement said, we consider character $\chi$ of $\theta_{i,\rho}$
for $a\in A$. Recall the character for induced representation for $s\in G$,
$H$ subgroup of $G$, $f:H\to \mathbf{C}^{\times}$, is
$$\text{Ind}_H^G(f)(s)=f'(s)=\frac{1}{|H|}\sum_{t\in G, t^{-1}st\in H}f(t^{-1}st)$$
Hence,
\begin{align*}
\chi(a) & = \frac{1}{|G_i|}\sum_{t\in G, t^{-1}at\in G_i}
\chi_{\chi_i\otimes \tilde{\rho}}(t^{-1}at), \\
& = \frac{1}{|G_i|}\sum_{t\in G} \chi_i(t^{-1}at)\chi_{\tilde{\rho}}(t^{-1}at),
\; (A \text{ normal so } t^{-1}at\in A\subset G_i \; \forall t\in G), \\
& = \frac{\text{dim }V}{|G_i|}\sum_{t\in G}\chi_i(t^{-1}at), \;
(t^{-1}at\in A \Rightarrow \tilde{\rho}(t^{-1}at)=\rho(1)=\text{id}_V), \\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H}\chi_i((a'h)^{-1}a(a'h)),\\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H} \chi_i(h^{-1}ah), \;
(A \text{ abelian}) \\
& = \frac{|A|\text{dim }V}{|G_i|} \sum_{h\in H} (h\chi_i)(a).
\end{align*}
Note that $h\chi_i\in X=\text{Hom}(A,\mathbf{C}^{\times})$ is
an irreducible representation of $A$ so the above implies that
the restriction of $\theta_{i,\rho}$ to $A$ is the direct sum of
representations corresponding to $h\chi_i$. Note that $h\chi_i$'s lie
in the orbit $H\chi_i\subset X$, which is disjoint with $H\chi_{i'}$
for $i\ne i'$. Hence, if $i\ne i'$, the restriction of $\theta_{i,\rho}$
and $\theta_{i',\rho'}$ to $A$ are not isomorphic.
(Serre) Let $W$ be the representation space for $\theta_{i,\rho}$, let
$W_i$ be the subspace of $W$ corresponding to $\chi_i$ [the set
of $x\in W$ such that $\theta_{i,\rho}(a)x=\chi_i(a)x$ for all
$a\in A$]. The subspace $W_i$ is stable under $H_i$ and the representation
of $H_i$ in $W_i$ is isomorphic to $\rho$; whence $\theta_{i,\rho}$
determines $\rho$.
Recall from $\S 7.1$ of Serre's book, $W$ can be identified with
$W=\mathbf{C}[G]\otimes_{\mathbf{C}[G_i]}V$ where
$V$ is the representation space of $\chi_i\otimes \tilde{\rho}$.
In particular, this is a $\mathbf{C}[G]$-module where $g$ acts
on $g'\otimes v$ by $(gg')\otimes v$. Also note that,
$(gg_i)\otimes v$ and $g\otimes (\chi_i\otimes \tilde{\rho})(g_i)v$
are considered the same in $W$ for $g_i\in G_i$.
Now, we identify $W_i$. For $a\in A, g\otimes v\in W$, we have
$\theta_{i,\rho}(a)(g\otimes v)= (ag)\otimes v
= g \otimes (g^{-1}ag)v$. Since $g^{-1}ag\in A$ so
$(\chi_i\otimes \tilde{\rho})(g^{-1}ag)v=\chi_i(g^{-1}ag)v$.
Thus, $\theta_{i,\rho}(a)(g\otimes v)= \chi_i(g^{-1}ag)
(g\otimes v)$. Thus, in order for
$\theta_{i,\rho}(a)(g\otimes v)=\chi_i(a)(g\otimes v)$ for all
$a\in A$, we must have $\chi_i(a)=(g\chi_i)(a)$ for all $a\in A$.
This follows $g\in H_i$. Thus, $W_i=\mathbf{C}[H_i]
\otimes_{\mathbf{C}[G_i]} V$.
With this, it is obvious that $W_i$ is stable under $H_i$.
Furthermore, observe $W_i$ is spanned by $1\otimes v_j$
where $v_j$'s basis of $V$, $1$ identity in $H_i$ (this holds
since $h\otimes v=1\otimes \rho(h)v$ for all $h\in H_i,v\in V$).
Hence, one can easily construct isomorphism representation of
$H_i$ in $W$ to $\rho$, as desired.
Best Answer
The proof is rather simple, just calculate characters (for rotations around OZ, since the axis does not affect the character) , and, using orthogonality theorem, note that all Fourier series coefficient for any other character are zero. But functions $\cos (l*\phi) (l - n)$, (where $n$ is an integer) form a complete set on $<0, \pi>$, so there are no more irreducible, unequivocal representations of $SO(3)$.