A good way to see this is to think of the Weyl fields as part of a Dirac field \(\begin{pmatrix}ψ_L \\ ψ_R \end{pmatrix}\). Then the matrix \(π\)is actually the off-diagonal component of a more general coupling:

\(\begin{pmatrix}\bar ψ_L & \bar ψ_R \end{pmatrix}\begin{pmatrix}0 & π \\ π^\dagger & 0 \end{pmatrix} \begin{pmatrix}ψ_L \\ ψ_R \end{pmatrix}\)

The symmetry transformations are part of a larger matrix

\(U = \begin{pmatrix}L & 0 \\ 0 & R \end{pmatrix}\)

and now the coupling transforms as \(UAU^\dagger\)as you would expect.

But even if the situation could not be explained as neatly as above: if the Lagrangian is symmetric under a fancy transformation, then this transformation is a "good" transformation, regardless of whether it fits a preconceived notion of what symmetry transformations should look like.