Will orientation-preserving diffeomorphisms of the plane preserve "positively oriented" domains? Background: Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be an orientation-preserving diffeomorphism, i.e. $\det(T')$ is everywhere positive in $\mathbb{R}^2$. Let $\gamma:[a,b]\rightarrow \mathbb{R}^2$ be a simple closed smooth regular plane curve such that it is positively oriented: the bounded component of $\mathbb{R}^2\backslash\gamma([a,b])$, call it D, lies on the left-hand side of $\gamma.$ More rigorously, $\forall t\in[a,b], \exists \epsilon=\epsilon(t)>0$ such that $\gamma(t)+i\cdot\epsilon(t) \dot\gamma(t)\in D$. Similarly we can define orientation-reversing diffeomorphisms and negatively oriented domains.
Question: $T^{-1}\circ\gamma$ be positively oriented with respect to $T^{-1}(D)$? Moreover, if $T$ is orientation-reversing ($\det(T')<0$ everywhere), will $T^{-1}\circ\gamma$ be negatively oriented with respect to $T^{-1}(D)$? 
Motivation: This fact is used in a proof of change of variable formula (for double integrals) via Green's theorem which requires the domain of integration to be positively oriented. For details see Exercise 15 of Section 7, Chapter 1 of Manfredo do Carmo's differential geometry book.
Edit:
$i\cdot\epsilon(t)\dot{\gamma}(t)$ means 
$\begin{pmatrix}0&-1\\ 1& 0\end{pmatrix}\epsilon(t)\dot{\gamma}(t)$
 A: Yes. The key fact in 2D is that for any $2\times 2$ matrix $M$, the product $i^T Mi$ is the cofactor matrix of $M$, or in other words when $M$ is invertible,
$$i^T M i = (\det M) M^{-T}.$$
This can be verified by direct computation:
$$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix} a & b \\ c& d\end{bmatrix}\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} c & d \\ -a& -b\end{bmatrix}\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix}d & -c \\ -b & a\end{bmatrix}.$$
Now to your question: let's work with $S=T^{-1}$ (which is orientation-preserving iff $T$ is) and write $\psi = S \circ \gamma.$
Then the pushforward of the original normal vector is $n=S'i\gamma'$ and the normal vector to $\psi$ is
$$m=i\psi' = iS'\gamma'.$$
We then have
\begin{equation}
\langle n, m\rangle = [\gamma']^Ti^T [S']^TiS'\gamma' = [\gamma']^T \det S' [S']^{-1}S'\gamma' = (\det S') \|\gamma'\|^2, \tag{*}
\end{equation}
and $S$ flips preserves the normal direction if $\det S' > 0$ and reverses it if $\det S'<0$.
The above calculation (*) generalizes to higher dimensions where $\gamma$ is a hypersurface of codimension one, and the Hodge star plays the role of the '$i$' operator.
