Smooth diffeomorphisms preserving common symmetries Let $A$ and $B$ be two $C^\infty$ orientation-preserving diffeomorphic, connected,  bounded, open subsets of ${\mathbb R}^n$, with finitely many ends and $G$ the group of isometries of ${\mathbb R}^n$ that preserve both $A$ and $B$ $\big(\phi (A)=A$ and $\phi (B)=B$, for any $\phi\in G \big)$.
Question: is there always an orientation-preserving $C^\infty$ diffeomorphism $f$ from $A$ to $B$ such that for every $K\subset A$ preserved by $G$, $f(K)$ is also preserved by $G$ i.e., $$\phi (K)=K, \; \forall \phi\in G \implies\phi\big(f(K)\big)=f(K), \; \forall \phi\in G\;?$$
i.e. an orientation-preserving smooth diffeomorphism that "preserves all the symmetries common to $A$ and $B$"? 
Partial answers in dimensions 2 or 3 will be also greatly appreciated.
Edit: counterexample in $\mathbb{R}^3$, essentially following user studiosus idea:
in the plane, let $\varLambda$ be the minimal invariant set under the $+\pi/2$ rotation around the origin $O_2$, that contains $ \{(1-1/n,0): n\geq 2\}$ ("the symbol "+" made of dots, with the center removed..."). Take $(\mathbb{B}^2\setminus \varLambda)\times ]-1,1[$ as $A$ and $A$ less $ \{O_2\} \times ]-1,1[$  as $B$. Then $A$ and $B$ are (orientation-preserving) smoothly diffeomorphic, with only one end and have the same group $G$ of symmetries in $\mathbb{R}^3$. $K=\{O_3\}\subset A$ is fixed by $G$ and in  particular by the $+\pi/2$ rotation $\phi$ around the $z$-axis. But no point of $B$ is fixed by $\phi$. Hence the desired diffeomorphism/homeomorphism does not exist.
Edit: a natural question arises: to characterize the open sets for which the conclusion above holds. In particular, is it valid when $A$ and $B$ are homeomorphic (and thus $C^\infty$ diffeomorphic) to $\mathbb{B}^2$ with finitely many points removed?
 A: The answer is no already in the plane. Take G of order 2 fixing the origin and A is the unit disk invariant under G with infinitely many punctures excluding the origin while B is a similar disk with infinitely many punctures including the origin. 
In order to handle the new condition on the number if ends, take this example and multiply both domains by the open unit interval. The result is a 3d example. 
Incidentally, I am not sure what the answer is if you add the convexity requirement for both domains. I suspect that contractibility is not enough.
Edit 1: Indeed, there are contractible examples in high dimension. This can be proven using the example discussed in 
Homeomorphic to the disk implies existence of fixed point common to all isometries? which has to be embedded equivariantly in some Euclidean space so that the Riemannian isometry group maps to a group of Euclidean isometries. Writing details is a bit too painful. 
Edit 2: For convex and, more generally, starlike, domains the answer to the question is positive. Here is the proof in the convex case. Let $U$ be a bounded convex domain, it has a well defined center of mass $c$ which therefore has to be fixed by the full group isometries $G$ of $U$. Then there exists a $G$-equivariant diffeomorphism from a small open ball centered at $c$ to $U$. Applying this to a pair of convex domains $U, V$ we obtain a $G$-equivariant diffeomorphism between the domains.
In the starlike case: If you have a starlike case you define the set of centers of a starlike domain $U$ to be the set of all points in $U$ with respect to which $U$ is starlike. Then the set of centers is open, convex and invariant under the full group of isometries of $U$. This reduces the problem to the convex case. 
