Let $\tau^\prime=a^{-1}(\tau)$. Then the linear transformation $$ \binom{z_1}{z_2}\mapsto j(a^{-1},\tau)^{-1}a \binom{z_1}{z_2} $$ tranforms the lattice $\cal O\binom{\tau^\prime}{1}$ into $a\cal Oa^{-1}\binom{\tau}{1}$.

Let $\tau^\prime=a^{-1}(\tau)$. Then the linear transformation $$ \binom{z_1}{z_2}\mapsto j(a^{-1},\tau)^{-1}a \binom{z_1}{z_2} $$ tranforms the lattice $\cal O\binom{\tau^\prime}{1}$ into $a\cal Oa^{-1}\binom{\tau}{1}$.

You shouldn't expect to realize the isomorphism with $\tau^\prime=\tau$. To identify $\cal O\otimes\Bbb R$ with $\Bbb C^2$ via action on $\binom\tau1$ amounts to endow $\cal O\otimes\Bbb R$ with a complex structure. In general conjugation on a real space doesn't preserve a complex structure.