= Solution
Set $\hbar=1$ and let $N=2I\in\mathbb Z_{\geq0}$. A concrete <irreducible spin representation> is the <symmetric power>
$$
\mathcal H_I=\operatorname{Sym}^{N}\mathbb C^2.
$$
Start with $N$ copies of the defining <SU(2)> doublet and restrict their <tensor product representation> to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\ldots,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i=\sum_{a=1}^N\sigma_i^{(a)}/2$, acting on this subspace, where $\sigma_i$ are the <Pauli matrices>. This also constructs the trivial representation when $I=0$.
The <spin ladder operators> $J_\pm=J_1\pm iJ_2$ satisfy
$$
[J_3,J_\pm]=\pm J_\pm,\qquad [J_+,J_-]=2J_3.
$$
The <Casimir operator> $J^2=\sum_iJ_i^2$ commutes with every generator. On the <highest-weight vector> $|I,I\rangle$, the identity $J^2=J_-J_++J_3(J_3+1)$ gives its <eigenvalue> $I(I+1)$. With phases chosen to make the lowering coefficients positive,
$$
\boxed{J_-|I,m\rangle=\sqrt{(I+m)(I-m+1)}\,|I,m-1\rangle,}
$$
$$
J_+|I,m\rangle=\sqrt{(I-m)(I+m+1)}\,|I,m+1\rangle.
$$
The squared <norm> of the lowered state follows from $J_+J_-=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct <eigenvalues>. Hence any <invariant subspace> contains a weight vector and then the whole ladder: the representation is irreducible.
For $k=I-m$, successive lowering has squared <norm>
$$
\prod_{r=0}^{k-1}(2I-r)(r+1)=\frac{(2I)!k!}{(2I-k)!}.
$$
Thus the <normalized highest-weight lowering formula> is
$$
\boxed{|I,m\rangle=\sqrt{\frac{(I+m)!}{(2I)!(I-m)!}}\,(J_-)^{I-m}|I,I\rangle.}
$$
All factorial arguments are nonnegative integers, including for half-integral $I$.
The <unitary representation> of an <isospin rotation> is
$$
\boxed{U[R(\theta,\mathbf n)]=e^{-i\theta\mathbf n\cdot\mathbf J}.}
$$
Its <matrix> in the weight basis is $D^{(I)}_{m'm}(R)=\langle I,m'|U[R]|I,m\rangle$. Insert the completeness relation between two operators to obtain
$$
D^{(I)}(g_1g_2)=D^{(I)}(g_1)D^{(I)}(g_2),\quad
D^{(I)}(1)=I,\quad D^{(I)}(g^{-1})=D^{(I)}(g)^\dagger.
$$
These verify the <group representation> and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their <SU(2)> lifts: $U(2\pi)=(-1)^{2I}I$. Integer $I$ descends to the ordinary <SO(3) group>; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.
For $I=1$, order the basis as $(|1,1\rangle,|1,0\rangle,|1,-1\rangle)$. The lowering and raising coefficients give the <spin-one half-turn matrix>
$$
J_2^{(1)}=\frac1{\sqrt2}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\qquad
(J_2^{(1)})^2=\frac12\begin{pmatrix}1&0&-1\\0&2&0\\-1&0&1\end{pmatrix}.
$$
Multiplication shows $(J_2^{(1)})^3=J_2^{(1)}$. Reduce the <matrix exponential> using this identity:
$$
e^{-i\theta J_2}=I-i\sin\theta\,J_2+(\cos\theta-1)J_2^2.
$$
At $\theta=\pi$ it becomes
$$
\boxed{e^{-i\pi J_2}=\begin{pmatrix}0&0&1\\0&-1&0\\1&0&0\end{pmatrix},\qquad
e^{-i\pi J_2}|1,m\rangle=-(-1)^m|1,-m\rangle.}
$$
Thus its <matrix> elements are $-(-1)^m\delta_{m',-m}$. In the <pion> phases $|\pi^\pm\rangle=|1,\pm1\rangle$, $|\pi^0\rangle=|1,0\rangle$, this <isospin rotation> exchanges the two charged states with positive coefficients and negates the neutral state.
<Charge conjugation> is linear and unitary here. Similarity preserves <commutators>, so
$$
[\mathcal CJ_i\mathcal C^{-1},\mathcal CJ_j\mathcal C^{-1}]
=\mathcal C[J_i,J_j]\mathcal C^{-1}
=i\epsilon_{ijk}\mathcal CJ_k\mathcal C^{-1}.
$$
Consequently the conjugated generators obey the same <Lie algebra>. The specified signs also give $\mathcal CJ_\pm\mathcal C^{-1}=-J_\mp$. Since $|\pi^\pm\rangle=J_\pm|\pi^0\rangle/\sqrt2$ and the neutral state has positive <charge conjugation> <eigenvalue>,
$$
\boxed{\mathcal C|\pi^\pm\rangle=-|\pi^\mp\rangle.}
$$
The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.
Conjugation by $R_\pi=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as <charge conjugation> does. Their product, the <G parity> operator, therefore satisfies
$$
GJ_iG^{-1}=\mathcal C(R_\pi J_iR_\pi^{-1})\mathcal C^{-1}=J_i.
$$
The <Schur lemma> makes $G$ scalar on each irreducible <isospin> multiplet, hence independent of $m$. On the neutral <pion>, $G|\pi^0\rangle=-\mathcal C|\pi^0\rangle=-|\pi^0\rangle$. Therefore the <pion G-parity> is
$$
\boxed{G=-1\quad\text{on all three pion states}.}
$$
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