Symmetries of contractable subsets of $\Bbb R^n$ Let $K\subset\Bbb R^n$ be a non-empty compact subset of $\Bbb R^n$. A symmetry of $K$ is an isometry of $\Bbb R^n$ that fixes $K$ set-wise. Since $K$ is compact, there is always a point $x\in\Bbb R^n$ fixed by all symmetries.

Question: Can the following three things be true at the same time:

*

*$K$ is contractible.

*$x$ is the only point fixed by all symmetries of $K$.

*$x\not\in K$.


The following image shows that such a $K$ exists if we ignore one of the properties:

 A: I posted a refined/generalized question Does a compact contractible metric space have a point that is fixed by all isometries? and received an answer that contained all essential ingredients to provide an affirmative answer to this question. The related question Homeomorphic to the disk implies existence of fixed point common to all isometries? has an answer with further very interesting details.
There are two ingredients:

*

*there exists a group $G$ that acts without fixed points on the $n$-dimensional ball (for an appropriate $n>5$).

*the ball can be smoothly embedded into some Euclidean space $\Bbb R^m$ so that the action of $G$ is now by linear transformations and only fixes the origin (Mostow's embedding theorem).

If we choose $K$ to be this embedded ball and $x$ to be the origin, then this configuration satisfies all the conditions from my question.
References to the claims 1 and 2 can be found in the linked answers, except for the assertion that the linear transformations only fix the origin, which I found in the original paper "Equivariant Embeddings in Euclidean Space" by Mostow.
A: As I see, the answer is “no”. If you have the single fixed point $X$ not from your set, then the symmetry of your compact is a reflection to the line $l$ which contains $X$.
It is easy to see that you cannot have another reflection line. Because if you have it, then their composition — rotation, is also a symmetry of your compact. But if you have a rotation, then $X$ is a centre of the corresponding circle and it's easy to the сontradiction to contractability.
And because of contractibility defined line l must intersect K. And that would be another fixed point which is in K.
