For two path-connected homeomorphic spaces, is there a homeomorphism with a prescribed value at a point? Let $X$, $Y$ be homeomorphic path-connected topological spaces. Is is true that for any pair $x\in X$,  $y\in Y$ a homeomorphism can be chosen such that $h(x)=y$?
 A: No, for instance, $X=Y=[1,2]$ (in their standard topology) are homeomorphic (and path-connected), but there is no homeomorphism mapping $1$ to $\frac{3}{2}$.
A: I will rephrase this as: given a space $X$ and two points $x,y \in X$, under what conditions is there a homeomorphism $f: X \to X$ such that $f(x) = y$?
One nice class of examples for which this is true for any pair $x,y \in X$ are connected topological manifolds without boundary: second-countable Hausdorff spaces locally homeomorphic to $\Bbb R^n$. Here is a sketch of a proof.
First, prove that for the closed unit ball $D^n$ and a pair of interior points $x,y \in D^n$, there is always a homeomorphism $f: D^n \to D^n$, identity on the boundary, that takes $p_1$ to $p_2$. One way of doing this is to use the homeomorphism $h: \Bbb R^n \to \text{int } D^n$ given by $h(x) = x \cdot \frac{\|x||}{1+\|x\|}$; let $f'(x) = x-h^{-1}(p_1) + h^{-1}(p_2)$, and show that $hf'h^{-1}$ extends to a homeomorphism $f: D^n \to D^n$ that fixes the boundary and sends $p_1$ to $p_2$.
Consider the equivalence relation on $X$: $x \sim y$ if there is a homeomorphism $\varphi: X\to X$ with $\varphi(x) = y$. We show equivalence classes are open. Pick a point $x \in X$; then it has a neighborhood $V$ homeomorphic to $\Bbb D^n$, with $x$ in the interior. There is a homeomorphim $D^n \to D^n$ that sends $x$ to any other interior point and fixes the boundary of $D^n$; extend this to a homeomorphism $X \to X$ (just make it the identity outside of $D^n$). So we see that on an open neighborhood $U = \text{int } V$ has that if $y \in U$, $x \sim y$; because this did not depend on $x$ we see that equivalence classes are open. If there were more than one equivalence class, this would provide a decomposition $X = U \cup U'$, where $U$ and $U'$ are disjoint and open. This contradict's $X$'s connectedness, so all points are equivalent. This is what we wanted.
The above is equivalent to saying that for a topological manifold $X$, the group $\text{Homeo}(X)$ acts transitively on $X$. If you set $\text{dim}(X) > 1$, more is true: $\text{Homeo}(X)$ acts $n$-transitively on $X$ for any $n$.
