= Solution
Use the <kinetic-energy inner product> $\langle v,u\rangle=\int v^*\cdot u\,dV$, with perturbation energy $\|u\|^2/2$. On divergence-free fields the primal generator, before the pressure projection, is
$$
A(t)u=-(\boldsymbol U\cdot\nabla)u-(u\cdot\nabla)\boldsymbol U+\mathrm{Re}^{-1}\nabla^2u.
$$
The base state is also divergence-free. The periodic boundary terms in $x,z$ cancel, and decay at $|y|\to\infty$ removes the remaining boundary terms. <Integration by parts> changes $-\boldsymbol U\cdot\nabla$ to $+\boldsymbol U\cdot\nabla$; the <Laplacian> is self-adjoint under these conditions. Writing $J_{ij}=\partial_jU_i$, the shear term $-Ju$ has <adjoint operator> $-J^Tv$. A <pressure gradient> pairs to zero with a divergence-free field. Hence
$$
A^\dagger(t)v=(\boldsymbol U\cdot\nabla)v-(\nabla\boldsymbol U)^Tv+\mathrm{Re}^{-1}\nabla^2v-\nabla\pi,
$$
where the last term enforces the divergence-free projection.
To conserve $\langle v(t),u_p(t)\rangle$ over the finite interval, the physical-time adjoint obeys $-\partial_tv=A^\dagger(t)v$. Put $\tau=-t$ and $u_d(\tau)=v(-\tau)$; then $\partial_\tau u_d=A^\dagger(-\tau)u_d$. This reverses the interval from $t=T$ towards $t=0$, or $\tau=-T$ towards $\tau=0$.
For $\boldsymbol\Omega=\nabla\times\boldsymbol U$, the <vorticity cross-product identity> and the <curl of a cross product>
$$
\boldsymbol\Omega\times u_d=(u_d\cdot\nabla)\boldsymbol U-(\nabla\boldsymbol U)^Tu_d,\qquad
\nabla\times(\boldsymbol U\times u_d)=(u_d\cdot\nabla)\boldsymbol U-(\boldsymbol U\cdot\nabla)u_d
$$
hold when both fields are divergence-free. Their difference equals the transport and shear terms in the <adjoint operator>. Therefore \b[the adjoint linearized Navier-Stokes evolution is]
$$
\boxed{\partial_\tau\boldsymbol u_d=\boldsymbol\Omega(-\tau)\times\boldsymbol u_d-
\nabla\times[\boldsymbol U(-\tau)\times\boldsymbol u_d]-\nabla p_d+\mathrm{Re}^{-1}\nabla^2\boldsymbol u_d,\qquad
\nabla\cdot\boldsymbol u_d=0.}
$$
The <Leray-Helmholtz projection> determines the adjoint pressure. The time-dependent base state is evaluated at $-\tau$; no extra time derivative of that base state appears in the spatial <adjoint operator>.
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