= Solution
\b[Angular momentum and parity]. Work in the <rho meson> <rest frame>. Since each <pion> has zero <spin>, their combined intrinsic <spin> is $S=0$ and their total <angular momentum> is entirely relative <orbital angular momentum>:
$$
\mathbf J_\rho=\mathbf L+\mathbf S_1+\mathbf S_2=\mathbf L,
\qquad J_\rho=1\ \Longrightarrow\ L=1.
$$
The <parity selection rule for a two-body decay> gives
$$
P_\rho=P_{\pi}^2(-1)^L=(-1)^2(-1)^L=(-1)^L.
$$
The supplied $P_\rho=-1$ requires odd $L$, so both conservation laws agree:
$$
\boxed{L=1,\qquad P_{\pi\pi}=-1.}
$$
Here the statement about two-pion decays is understood as the dominant <strong interaction> channel; the assertion that it occurs literally always is an idealization, since other interactions can produce other channels.
\b[Exchange symmetry]. <Pions> are spin-zero <bosons>, so the complete two-particle <wavefunction> must have <bosonic exchange symmetry>, including its spatial and <isospin> factors. The unique spin-zero product is symmetric. Exchanging the particle slots reverses the relative position, so an $L=1$ spatial state has <orbital parity> $(-1)^L=-1$. Its <isospin> state must therefore be antisymmetric to make the product symmetric.
For two <neutral pions>, the internal charge state $|\pi^0\pi^0\rangle$ is symmetric under exchange. There is no antisymmetric internal factor to compensate odd $L$. Equivalently, the odd-$L$ projection of two identical spin-zero neutral particles vanishes. Therefore
$$
\boxed{\rho\to\pi^0\pi^0\ \text{is forbidden by Bose exchange symmetry}.}
$$
This argument does not depend on treating <isospin> as an exact symmetry of nature.
\b[All nine isospin states]. In the following <two-pion isospin decomposition>, let $|ab\rangle=|\pi^a\rangle_1\otimes|\pi^b\rangle_2$ denote ordered particle slots; each of $a,b$ is $+,0,-$. The second label in $|I,q\rangle$ is the total <electric charge> in units of the <elementary charge>. For this pion system it equals the total isospin projection $I_3$.
<Addition of angular momentum> for two <isospins> of one gives $1\otimes1=2\oplus1\oplus0$. Start from the highest-weight <quantum state> $|2,2\rangle=|++\rangle$ and apply the sum of the two <ladder operators>. The <ladder recurrence for Clebsch-Gordan coefficients> gives the normalized symmetric quintet:
$$
\begin{aligned}
|2,2\rangle&=|++\rangle,\\
|2,1\rangle&=\frac{|+0\rangle+|0+\rangle}{\sqrt2},\\
|2,0\rangle&=\frac{|+-\rangle+|-+\rangle+2|00\rangle}{\sqrt6},\\
|2,-1\rangle&=\frac{|0-\rangle+|-0\rangle}{\sqrt2},\\
|2,-2\rangle&=|--\rangle.
\end{aligned}
$$
The normalized <quantum state> orthogonal to $|2,1\rangle$ in the two-dimensional charge-one subspace is antisymmetric. Lowering it gives the antisymmetric triplet:
$$
\begin{aligned}
|1,1\rangle&=\frac{|+0\rangle-|0+\rangle}{\sqrt2},\\
|1,0\rangle&=\frac{|+-\rangle-|-+\rangle}{\sqrt2},\\
|1,-1\rangle&=\frac{|0-\rangle-|-0\rangle}{\sqrt2}.
\end{aligned}
$$
The remaining charge-zero <quantum state> must be orthogonal to both displayed charge-zero states and annihilated by the total <ladder operators>. This gives the symmetric singlet:
$$
\boxed{|0,0\rangle=\frac{|+-\rangle+|-+\rangle-|00\rangle}{\sqrt3}.}
$$
Together the quintet, triplet and singlet contain $5+3+1=9$ orthonormal states, accounting for the full $3\times3$ charge-product space. Multiplying an entire <isospin multiplet> by a common <quantum phase> changes none of these physical conclusions.
\b[Rho isospin and strong decay modes]. Only the $I=1$ <two-pion isospin decomposition> is antisymmetric under exchange; $I=0,2$ are symmetric. The required $L=1$ spatial state therefore selects $I_{\pi\pi}=1$. Conservation of <isospin> by the <strong interaction>, in its usual approximation, implies
$$
\boxed{I_\rho=1.}
$$
The three charge states have the corresponding strong two-pion channels
$$
\boxed{\rho^+\to\pi^+\pi^0,\qquad
\rho^0\to\pi^+\pi^-,\qquad
\rho^-\to\pi^-\pi^0.}
$$
The ordered antisymmetric combinations above describe the internal part; multiplying by the odd spatial state gives the required symmetric complete <bosonic> state. The neutral $I=1$ state contains no $|00\rangle$ component, in agreement with the independent identical-particle argument.
\b[Charge conjugation]. <Charge conjugation> preserves <mass> and reverses <electric charge>. A charged <rho meson> is therefore changed into a distinct state of opposite charge, so neither $\rho^+$ nor $\rho^-$ is itself a charge-conjugation eigenstate. The neutral state can be an eigenstate. Acting on its two-pion internal state interchanges the charge labels:
$$
\mathcal C\frac{|+-\rangle-|-+\rangle}{\sqrt2}
=\frac{|-+\rangle-|+-\rangle}{\sqrt2}
=-\frac{|+-\rangle-|-+\rangle}{\sqrt2}.
$$
The charged-pion <quantum phase> conventions cancel between the two particles. Thus the <two-pion charge-conjugation parity> is $C=(-1)^L=-1$, and conservation by the <strong interaction> fixes
$$
\boxed{\mathcal C|\rho^0\rangle=-|\rho^0\rangle;\qquad C(\rho^0)=-1.}
$$
Only $\rho^0$ among the individual charge states is a <charge conjugation> eigenstate.
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