Let $S$ be a (unit) sphere in a Hilbert Space $H$ with $\dim H \ge 3$. Let $A \subset S$ have the following properties:


1. $A$ is connected;
2. The affine hull of $A$ is the whole space;
3. For every $x,y\in A$ there is a unitary operator $T:H\to H$ (or, alternatively, an isometry $T:S\to S$), such that $TA=A$, and such that $Tx=y$.


Is it true that $A=S$?


You can assume that $A$ with the induced strong or weak topology is locally compact and locally linearly connected if needed.


I think that the problem may be meaningful even in dimension $3$, where it would be nice to have an elementary solution.


The problem I have started with is slightly more general, in case of the affirmative answer for this one, there may be a follow-up.


Thank you.



Homogeneous subsets of the sphere
The answer is 'no'. 


For example, let $H$ be the (5-dimensional) space of symmetric, traceless $3$-by-$3$ matrices, where the Hilbert inner product is $\langle x,y\rangle = \tfrac12\mathrm{tr}(xy)$. Let $A\subset S$ be the subset consisting of those matrices of unit norm for which $0$ is an eigenvalue. Then $A$ is connected and $\mathrm{SO}(3)$ acts transitively on $A$ by the action $g\cdot x = gxg^T$. Moreover, the affine hull of $A$ must be a linear subspace of $H$ containing $A$ that is invariant under the action of $\mathrm{SO}(3)$, and this implies that this affine hull is all of $H$.


More generally, if $G$ is a connected compact Lie group acting irreducibly by isometries on a (finite-dimensional) Hilbert space $H$, then each of the orbits of $G$ in the unit sphere $S\subset H$ will be a subset $A\subset S$ with the OP's desired properties, and such orbits will not be all of $S$ whenever $G$ does not act transitively on $S$.


