= Solution
First choose the same orthonormal <basis> for the left and right copies of the two-dimensional <unitary representation> $D$. Then the family <representation> is $F=D\oplus\varepsilon$. Because the barred <fermion> transforms with $F(g)^\dagger$ and the <scalar field> $H$ is a <trivial representation>, invariance of the <Yukawa interaction> requires
$$
F(g)^\dagger\lambda F(g)=\lambda,
\quad\text{equivalently}\quad \lambda F(g)=F(g)\lambda
\qquad(g\in S_3).
$$
Write $\lambda$ in a $2+1$ block form. The off-diagonal blocks would be <intertwiners> between inequivalent <irreducible representations> $D$ and $\varepsilon$, so the <Schur lemma> makes them zero. The doublet block must be a scalar <matrix>, while the singlet block is arbitrary. Therefore
$$
\boxed{\lambda=\begin{pmatrix}a&0&0\\0&a&0\\0&0&b\end{pmatrix},\qquad a,b\in\mathbb C.}
$$
Equivalently, the allowed <Yukawa interaction> is $H[a(\bar\psi_L^1\psi_R^1+\bar\psi_L^2\psi_R^2)+b\bar\psi_L^3\psi_R^3]$, together with its <Hermitian conjugate> when constructing a real <Lagrangian>. The <tensor product> rules explain the two free parameters: both $D^*\otimes D$ and $\varepsilon^*\otimes\varepsilon$ contain one <trivial representation>, while the mixed products contain none.
There is a useful <basis> qualification. The <basis> used in <2/c/solution> is not orthonormal: its <Gram matrix> is $K=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, and $D(g)^\dagger KD(g)=K$. If the fields are expressed in exactly those coordinates, with the barred components defined by ordinary conjugate transpose, the invariant doublet coefficient is $aK$, rather than $aI_2$. Canonically normalizing the <kinetic term> returns the boxed form. Independently chosen left and right orthonormal <bases> replace $aI_2$ by $aS$, where $S$ is their fixed unitary <intertwiner>; the physical degeneracy is unchanged.
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