This question is inspired by this one, or rather the way I understood it.
Let $\gamma$ and $\delta$ be parametrised Jordan curves on the plane (i.e. homeomorphisms from $S^1$ onto its image in $\mathbf{R}^2$).
For $s\in S^1$ define a (Jordan) curve $\gamma_{s}$ by $\gamma_{s}(t)=\gamma(t)-\gamma(s)+\delta(s)$. This is just a translation of $\gamma$, so that its value at $s$ was the same as of $\delta$.
Assume that $\gamma_{s}(S^1)\cap\delta(S^1)=\{\delta(s)\}$, for every $s$. Geometrically, this means that we slide (the image of) $\gamma$ on (the image of) $\delta$ by translating it, and $\gamma$ envelopes $\delta$.
Does it follow that there are convex bodies $G$ and $D$ in $\mathbf{R}^2$, such that $\gamma(S^1)=\partial G$, $\delta(S^1)=\partial D$ and there are parallel support lines to $G$ at $\gamma(s)$ and to $D$ at $\delta(s)$, for every $s$?
Note that in the original question $\delta=r\circ\gamma$, where $r$ is a rotation in $\frac{\pi}{2}$, and the condition $\gamma_{s}(S^1)\cap\delta(S^1)=\{\delta(s)\}$ was satisfied not for every $s\in S^1$, but for every $s$ in a dense subset of $S^1$ (if I understand that question correctly).
If both $\gamma$ and $\delta$ are smooth at $s$, the condition means that at least the tangents to these curves at the corresponding points are parallel, but I have no idea what happens if the curves are nasty.
