TITLE: Infinite vertex-transitive graph where every automorphism has a fixed vertex
QUESTION [11 upvotes]: This is a follow-up to the question Connected vertex-transitive graph with the fixed-point property. In particular, it is based on a comment by user bof.
Let $G = (V,E)$ be a graph with $V$ infinite. Suppose $G$ is vertex-transitive, i.e., for every pair of vertices $u,v \in V$ there is an automorphism $\gamma$ of $G$ for which $\gamma(u) = v$. Is it possible that every automorphism of $G$ has at least one fixed vertex?
The previous question dealt with $V$ finite, for which the answer is no (as long as $\#V \geq 2$) by a simple group theory argument. But here it does not seem possible to use a group theory argument-- at least not so easily-- because there are transitive actions on infinite sets in which every permutation has a fixed point.

REPLY [6 votes]: Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^n$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.
Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.