= Solution
Use a finite-dimensional <unitary representation> of <SU(2)> and units $\hbar=1$, so the <isospin> generators are <Hermitian> and $I_+^\dagger=I_-$. Their commutators are
$$
[I_3,I_\pm]=\pm I_\pm,\qquad[I_+,I_-]=2I_3.
$$
On the normalized <highest-weight vector>, the given <quadratic Casimir operator> identity gives $\mathbf I^2|I,I\rangle=I(I+1)|I,I\rangle$. Since the <quadratic Casimir operator> commutes with every generator, this eigenvalue is unchanged down the ladder. Each application of $I_-$ lowers $M$ by one. For a normalized descendant,
$$
\|I_-|I,M\rangle\|^2=\langle I,M|I_+I_-|I,M\rangle=I(I+1)-M^2+M=(I+M)(I-M+1).
$$
Positivity first gives $I\ge0$. In a finite-dimensional representation the descending ladder must terminate; the norm vanishes only at $M=-I$. Thus the number of lowering steps is $2I$, a nonnegative integer, and
$$
\boxed{I\in\{0,\tfrac12,1,\tfrac32,\ldots\},\qquad I_-^{2I+1}|I,I\rangle=0.}
$$
For $M>-I$, define the normalized descendant by dividing $I_-|I,M\rangle$ by its positive norm. This fixes its relative phase and gives
$$
I_-|I,M\rangle=\sqrt{(I+M)(I-M+1)}|I,M-1\rangle.
$$
Acting with $I_+$ and using $I_+I_-=\mathbf I^2-I_3^2+I_3$ proves
$$
I_+|I,M-1\rangle=\sqrt{(I+M)(I-M+1)}|I,M\rangle.
$$
The resulting <isospin multiplet> has $2I+1$ states. More explicitly,
$$
|I,M\rangle=\sqrt{\frac{(I+M)!}{(2I)!(I-M)!}}\,I_-^{I-M}|I,I\rangle.
$$
The factorial arguments are integers, even for half-integer $I$.
Put $R(\theta)=e^{-i\theta I_2}$. Differentiation and the commutators give
$$
R(\theta)I_1R(\theta)^{-1}=I_1\cos\theta-I_3\sin\theta,\qquad R(\theta)I_3R(\theta)^{-1}=I_3\cos\theta+I_1\sin\theta,
$$
while $I_2$ is unchanged. Hence at the half-turn,
$$
\boxed{RI_1R^{-1}=-I_1,\quad RI_2R^{-1}=I_2,\quad RI_3R^{-1}=-I_3,\qquad R=e^{-i\pi I_2}.}
$$
The half-turn preserves $I$ and reverses $M$, so $R|I,M\rangle=\alpha_M|I,-M\rangle$. Since $RI_+R^{-1}=-I_-$, comparison of the two ways to rotate a raised state gives $\alpha_{M+1}=-\alpha_M$.
To fix the remaining phase, expand the supplied highest-state rotation identity as a finite sum. The $k$th term is
$$
\frac{(\cos(\theta/2))^{2I-k}(\sin(\theta/2))^k}{k!}I_-^k|I,I\rangle,\qquad0\le k\le2I.
$$
As $\theta\uparrow\pi$, all terms vanish except $k=2I$. The product of the lowering coefficients gives $I_-^{2I}|I,I\rangle=(2I)!|I,-I\rangle$. The surviving coefficient is therefore one, fixing $\alpha_I=1$. The <half-turn phase of an SU(2) multiplet> is
$$
\boxed{\alpha_M=(-1)^{I-M}.}
$$
In particular the recurrence and highest-state phase are consistent with $R^2=(-1)^{2I}$.
<Charge conjugation> is a linear unitary symmetry. If $J_i=CI_iC^{-1}$, conjugation preserves the commutator and the scalar $i$, so
$$
[J_i,J_j]=C[I_i,I_j]C^{-1}=i\epsilon_{ijk}J_k.
$$
Thus the conjugated generators obey the same <SU(2) Lie algebra>. The <charge conjugation> action on $(I_1,I_2,I_3)$ is exactly the same pair of sign reversals as $R$. Their product $G=CR$ consequently satisfies $GI_iG^{-1}=I_i$ for all $i$. For integer $I$, its value on the neutral member is
$$
G|I,0\rangle=(-1)^I C|I,0\rangle=\eta(-1)^I|I,0\rangle.
$$
Commuting $G$ through the <ladder operators> transfers that value to every state in the same <isospin multiplet>. Therefore the <G-parity of an isospin multiplet> is
$$
\boxed{G|I,M\rangle=\eta(-1)^I|I,M\rangle.}
$$
A <photon> has <charge conjugation> value minus one, so a two-photon state has value plus one. The electromagnetic two-photon decay preserves <charge conjugation>, giving $\eta_{\pi^0}=+1$. Since the <pion> has $I=1$, all three <pion> charge states have <G parity> minus one:
$$
\boxed{G|\pi^{+,0,-}\rangle=-|\pi^{+,0,-}\rangle.}
$$
The <rho meson> has $I=1$, $\eta=-1$, hence $G_\rho=+1$, whereas the <omega meson> has $I=0$, $\eta=-1$, hence $G_\omega=-1$. On an $n$-pion state the internal symmetry acts on each <pion>, giving $(-1)^n$ independently of orbital <angular momentum>. Conservation of <G parity> in the isospin-symmetric <strong interaction> therefore gives the <G-parity selection rule for pion multiplicities>:
$$
\boxed{\rho\to2\pi\text{ allowed},\quad\omega\to3\pi\text{ allowed},\quad\rho\not\to3\pi,\quad\omega\not\to2\pi.}
$$
The allowed channels can also satisfy <angular momentum> and physical parity: the two <pions> from a rho can have relative orbital <angular momentum> one and total <isospin> one, while three <pions> can couple to the omega's $I=0$, $J^P=1^-$. The rule refers to the ideal isospin-conserving strong amplitude; small symmetry-breaking or electromagnetic contributions are not excluded.
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