Consider $A = \begin{bmatrix} 2 && 1 \\ 1 && 1 \end{bmatrix}$. Then $A$ acts on $\mathbb{T}^2 = (\mathbb{R}/\mathbb{Z})^2$ by matrix multiplication. If $f \in C(\mathbb{T}^2)$, then let $(\Phi_Af)(x) = f(Ax) $.
I want to compute a formal adjoint of $\Phi_A$, i.e. I want a map $L: C(\mathbb{T}^2) \to C(\mathbb{T}^2)$ with
$$ \int \Phi_A(f) g = \int f L(g)$$
I'm thinking you just change variables from $x$ to $A^{-1}x$ to conclude that $L(g)(x) = g(A^{-1}x)$. But, now, I'm worried about the exact domain of integration, since on the left I'm really integrating over $[0,1]^2$, while after the change of variables I get a different domain on the right. My gut tells me that because of the periodicity, the domain and domain change should not really matter. However, I'm having trouble exactly justifying why.
You may argue in different ways. You may do it by hand: take the unit square S and transform it (to a parallelogram). This you may chop it into 4 pieces and translate those back to precisely cover S. A more abstract approach (valid for general $A\in GL_n({\Bbb Z})$) is to prove that $f$ is a diffeomorphism that preserves Lebesgue measure. For this you need to show e.g. that $Ax-Ax'\in {\Bbb Z}^2$ iff $x-x'\in {\Bbb Z}^2$ which is due to the fact that $A$ and its inverse preserve the lattice.