Is there an area-preserving diffeomorphism of the disk which is nowhere conformal? This question is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the closed unit disk.

Does there exist a smooth area-preserving diffeomorphism $f:D \to D$ that does not have conformal points?

I want the singular values $\sigma_i$ of $df$ to be everywhere distinct, and $\det(df)=1$.

It is proven here that for any map $f:D \to \mathbb R^2$, the condition of being with distinct singular values is 'generic' in the following sense: There exist $f_n \in C^{\infty}(D, \mathbb{R}^2)$ such that $f_n \to f$ in $W^{1,2}(D, \mathbb{R}^2)$ and $df_n$ has distinct singular values everywhere on $D$.
It seems to me that this approximation procedure, applied to a map $f \in \text{Diff}(D)$, does not guarantee that the $f_n$ will map $D$ into $D$, let alone be diffeomorphisms. (e.g. I think that the convergence $f_n \to f$ cannot be made uniform in general). However, perhaps this genericity phenomena can still be used somehow.

This answer provides the following example for a one-parameter family of such diffeomorphisms $D\setminus \{0\} \to D \setminus \{0\}$:
$$f_c: (r,\theta)\to (r,\theta+c\log r).$$ This description is given in terms of polar coordinates both in the domain and the range. For each non-zero $c ֿ\in \mathbb R$ we get a diffeomorphism, with fixed distinct singular values whose product is $1$.

Edit:
Can we answer the infinitesimal version of the question? That is, let $f_t$ be a smooth family of area-preserving diffeomorphisms. Does each $f_t$ has a conformal point? This answer treats the "formally infinitesimal" case.
 A: Let me try to prove an infinitesimal version where an area preserving diffeomorphism is replaced by a Hamiltonian vector field.
Let H be a function in the unit disc, constant on the boundary. Consider the Hamiltonian vector field $(H_y,-H_x)$ and the respective infinitesimal diffeomorphism
$(x,y) \mapsto (x+\epsilon H_y, y-\epsilon H_x).$
When is it conformal at a point? When the Jacobian is a dilation. That is, if the matrix is
$\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}$,
then $a=d, b+c=0$. In our case, the Jacobian is
$\begin{pmatrix}
1+\epsilon H_{xy}& \epsilon H_{yy}\\
-\epsilon H_{xx}& 1-\epsilon H_{xy}.
\end{pmatrix}$.
Hence we want to find solutions of
$H_{xx}=H_{yy}, H_{xy}=0.$
Consider the vector field $V:=(H_{xx}-H_{yy}, 2H_{xy})$. If it has zero on the boundary, we are done. If not, let's calculate it in polar coordinates $(\alpha,r)$ on the boundary circle $r=1$. Chain rule calculations simplify by the fact that, on the boundary, $H_\alpha=0$. My calculation yields:
$V = H_{rr} (\cos 2\alpha, \sin 2\alpha).$
The index of this field equals 2, so there are two zeroes, multiplicities counted, inside the disc.
