Rigid space, but with homeomorphic neighborhoods

Question

What is an example of a topological space $(X,\tau)$ on more than one point, with the following properties?

1. the only homeomorphism from $X$ to itself is the identity, and
2. given $x,y\in X$ there are open sets $U, V$ with $x\in U, y\in V$ and a homeomorphism $\varphi: U\to V$ such that $\varphi(x) = y$, where $U,V$ are endowed with their respective subspace topologies.

Answer 1

For $n\in\mathbb N$ let $U_n=\{m\in\mathbb N:m\geq n\}$. Then $\tau=\{\varnothing\}\cup\{U_n:n\in\mathbb N\}$ is a topology on $\mathbb N$. This space is rigid because $n$ is characterized as the unique element contained in exactly $n+1$ open sets. However, for any $n,m$ the sets $U_n,U_m$ are homeomorphic through a map $k\mapsto k-n+m$ which takes $n$ to $m$.

Answer 2

There exists a metrizable topological group $H$ such that $H \setminus \{e\}$ is rigid (see Theorem 6.1 in van Mill´s paper: A topological group having no homeomorphisms other than translations).

Exercise: Without knowing anything else about $H$, show that $X=H \setminus\{e\}$ satisfies condition 2 in the OP.