This question is related to this one, though is supposed to be easier.
Let $D \subseteq \mathbb{R}^2$ be the closed unit disk.
Does there exist a smooth volume-preserving diffeomorphism $f:D \to D$ such that the singular values of $df$ are everywhere distinct?
i.e. I want $\sigma_1(df_p) \neq \sigma_2(df_p)$ for every $p \in D$, and the product $\sigma_1(df)\sigma_2(df)=1$ to be constant.
This answer provides the following example for such a diffeomorphism $D\setminus \{0\} \to D \setminus \{0\}$ with the required properties:
$f_c: (r,\theta)\to r\big(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))\big),\;\; $
(for every non-zero $c ֿ\in \mathbb R$ we get an example).
Edit-a description of a possible topological obstruction:
Set $\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$, where by a non-conformal matrix, I refer to a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).
Suppose that such an $f \in \text{Diff}(D)$ exists. Then $df|_{\partial D}:\partial D \to \mathcal{NC}$ is homotopic to a constant.
$df|_{\partial D}$ maps $T\partial D$ to itself, and in particular, at a specific point $\theta \in \mathbb{S}^1$, $(df|_{\partial D})_{\theta}(T_{\theta}\partial D)=T_{f(\theta)}\partial D$. So, thinking on $e_2$ as an element of $T_{(0,1)}\partial D$, we have
$R_{f(\theta)}^{-1} \circ df_{\theta} \circ R_{\theta}(e_2)=\lambda(\theta) e_2$, for some positive factor $\lambda(\theta)$.
Setting $A_{\theta}:=R_{f(\theta)}^{-1} \circ df_{\theta} \circ R_{\theta}$, and $$\mathcal{F}:=\{ A \in \text{SL}_2(\mathbb{R}) \, | \, Ae_2 \in \operatorname{span}(e_2) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\},$$ we get $$ df_{\theta} =R_{f(\theta)} \circ A_{\theta} \circ R_{-\theta}, \, \, \, A_{\theta}: \partial D \to \mathcal{F}. \tag{1}$$
If $\mathcal{F}$ were contractible in $\mathcal{NC}$, we could deform $A_{\theta}$ to a constant map $ \partial D \to \mathcal{NC}$.
Thus, by equation $(1)$, $df|_{\partial D}$ would be homotopic to the map $\theta \to R_{f(\theta)} \circ A \circ R_{-\theta}$ for some constant non-conformal matrix $A \in \mathcal{NC}$.
Writing $A=R_{\alpha} \Sigma R_{\beta}$ where $\Sigma$ is non-negative and diagonal, we would get that $df|_{\partial D}$ is homotopic to $\theta \to R_{f(\theta)+\alpha} \circ \Sigma \circ R_{-\theta+\beta}.$
On the space of non-conformal matrices $\mathcal{NC}$, there is a continuous map* $H:\mathcal{NC} \to \mathbb{S}^1$, given by $H(R_{\phi} \Sigma R_{\theta})= R_{2\theta}$. This leads to a contradiction to the contractibility of $df|_{\partial D}$:
Indeed, if it were homotopic to a constant, then so would the map $\theta \to R_{f(\theta)+\alpha} \circ \Sigma \circ R_{-\theta+\beta}.$ Composing it with $H$, we obtain the map $\theta \to R_{-2\theta+2\beta}$, or $\theta \to -2\theta$, which is not homotopic to a constant.
Since $\mathcal{F}$ is not contractible in $\mathcal{NC}$, this argument fails. However, perhaps a more refined topological argument could obtain more, I don't know.
*The map $H$ is well-defined, since $U\Sigma V^T=(-U)\Sigma (-V)^T$, and this is the only ambiguity in $U,V$ for a matrix in $\mathcal{NC}$. Thus $\theta$ is well defined up to an addition of $\pi$.