Homeomorphisms of an open set onto itself. Given two distinct points $a,b\in U$ where $U$ is an open subset of $\mathbb R^n$, can one always find a homeomorphism $f:U\to U$ with $f(a)=b$ or does one need any further conditions on $ U $ or the points? 
Thank you.
 A: I seems that all puzzle pieces to answer your question have been found in the 
above discussion. I'll try to summarize it:

Let $U$ be an open subset of $\mathbb R^n$ and $a, b \in U$ be
  distinct. A homeomorphism $f: U \to U$ with $f(a) = b$ exists
  if and only if the connected components $A, B$ of $U$ containing $a$
  and $b$ respectively, are homeomorphic.

In particular, such a homeomorphism exists if $a, b$ lie in the same
connected component of $U$.
The "only if" part is the easier one: If a homeomorphism $f: U \to U$ with $f(a) = b$ exists then $f(A) = B$ because homeomorphisms map connected components 
onto connected components, so $A$ and $B$ are homeomorphic.
For the "if" part assume that there is a homeomorphism $g: A \to B$. The crucial tool is
to use the result of 


*

*Are all connected manifolds homogeneous
which states that there is a homeomorphism $h: B \to B$ such that $h(g(a)) = b$. 
Then $F := h \circ g : A \to B$ is a homeomorphism from $A$ onto $B$ 
satisfying $F(a) = b$. $F$ can be extended to a homeomorphism $f:U \to U$:
In the case $A \ne B$ define
$$
\begin{align}
f(x) &= F(x)  &\text{ if } x \in A,  \, \\
f(x) &= F^{-1}(x) &\text{ if } x \in B,  \,\\
f(x) &= x \quad &\text{ if } x \in U \setminus (A \cup B) \, .
\end{align}
$$
$f$ "swaps" the components $A$ and $B$ and leaves all other components unchanged.
And if $A = B$ then define $f$ via
\begin{align}
f(x) &= F(x)  &\text{ if } x \in A,  \, \\
f(x) &= x \quad &\text{ if } x \in U \setminus A \, .
\end{align}
