Move one element of finite set out from A in plane

Suppose we are given two sets, $$S$$ and $$A$$ in the plane, such that $$S$$ is finite, with a special point, $$s_0$$, while neither $$A$$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $$A$$ and $$A^c=\mathbb R^2\setminus A$$ are both non-zero. Can we find two congruent copies of $$S$$, $$f_1(S)$$ and $$f_2(S)$$, such that $$f_1^{-1}(f_1(S)\cap A)\Delta f_2^{-1}(f_2(S)\cap A)=\{s_0\}$$, i.e., $$s_0$$ is the only element of $$S$$ that goes in to/out of $$A$$ when we go from $$S_1$$ to $$S_2$$?

My motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. It is easy to see that this can be done when the elements of $$S$$ are the three vertices of an equilateral triangle; in fact, the same trick works whenever $$S\setminus\{s_0\}$$ falls on a line, which doesn't contain $$s_0$$.

The problem seems quite similar to the finite Steinhaus problem/Jackson sets, as Laczkovich has pointed out.

• You reverted my tag edit, but this question has little to do with real analysis. – YCor Aug 15 '19 at 20:44
• @YCor Yes, because I think that many such problems are tagged real-anal, like mathoverflow.net/questions/219860, and I also had such problems on my real-anal class at the uni. While this is debatable, I'm sure that this question has nothing to do with discrete-geometry, which you've tagged it. – domotorp Aug 15 '19 at 20:55
• Agreed on discrete geometry. I don't think the linked post fits real-analysis either. On the other hand both clearly lack measure-theory. – YCor Aug 15 '19 at 21:02
• @YCor Yes, that's my main problem too, that if I ask a measure-theory type question about non-measurable sets, then it's hard to tag it. Btw, I've also tagged my previous, similar question real-anal: mathoverflow.net/questions/337847 – domotorp Aug 15 '19 at 21:05
• "Translate" in title, "congruent" in body. I think you mean "congruent". – bof Aug 15 '19 at 22:09