= Solution
Only the <singlet representation>[singlet] $|1\rangle$ and isoscalar <octet representation>[octet] member $|8_0\rangle$ can mix without violating <isospin>: they have the same $I=I_3=S=0$ and, for the vector multiplet, the same $J^{PC}=1^{--}$. The neutral <triplet representation>[triplet] member has $I=1$ and is excluded in the <isospin> limit.
Set $|\psi_a\rangle=|1\rangle$ and $|\psi_b\rangle=|8_0\rangle$, with precisely the phases in <3/a>. Define $|n\rangle=(|u\bar u\rangle+|d\bar d\rangle)/\sqrt2$. Taking overlaps gives
$$
|n\rangle=\sqrt{\frac23}|1\rangle-\frac1{\sqrt3}|8_0\rangle,
\qquad
|s\bar s\rangle=\frac1{\sqrt3}|1\rangle+\sqrt{\frac23}|8_0\rangle.
$$
Therefore the source's rotation convention is satisfied by
$$
\boxed{\cos\theta=\sqrt{\frac23},\qquad
\sin\theta=-\frac1{\sqrt3},\qquad
\theta=-\arctan\frac1{\sqrt2}\simeq-35.26^\circ.}
$$
This yields $|\omega\rangle=|n\rangle$ and $|\phi\rangle=|s\bar s\rangle$ exactly. Reversing the <octet representation>[octet] phase changes the quoted sign, with an irrelevant compensating phase on the physical phi state; exchanging the basis order gives the complementary angle. The physical statement of <ideal vector-meson mixing> is the pure strange and pure nonstrange content, rather than a convention-independent sign for the angle.
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