Let $G=(V,E)$ be an countably infinite, locally finite transitive graph. Say that $G$ is exchangeable if for every two vertices $v,w \in V$ there exists a graph homomorphism that maps $v$ to $w$ and $w$ to $v$. Certainly such a graph has a unimodular automorphism group. Is there an example of a graph $G$ that is not exchangeable but is unimodular?

1$\begingroup$ how is exchangeable different than vertex transitive? $\endgroup$ – Benjamin Steinberg Feb 10 '14 at 20:40

$\begingroup$ You are absolutely right  I left out the main point. Now edited to make sense. $\endgroup$ – Vladimir Feb 11 '14 at 1:32

$\begingroup$ There was a mistake in the initial phrasing of the problem. This mistake is now fixed. $\endgroup$ – Vladimir Feb 12 '14 at 14:06
The standard Cayley graph $\Gamma$ of the BaumslagSolitar group $BS(1,2) = \langle a,b \mid bab^{1} = a^2 \rangle$ is nonexchangable and unimodular (as is any Cayley graph). See http://en.wikipedia.org/wiki/BaumslagSolitar_group for a picture of $\Gamma$. To see nonexchangability, observe the following:
 an edge is labeled by $a$ iff it is contained in three distinct $5$cycles, and by $b$ iff it is contained in two distinct $5$cycles. Hence $\mathrm{Aut}(\Gamma)$ preserves the labeling of the (unoriented) edges by $a$ or $b$.
 any $5$cycles contains two $a$labeled components, one with two edges and one with a single edge. Given any $5$cycle, an oriented edge in the cycle is of the form $(g, gb)$ iff it points towards the $a$labeled component with a single edge. It follows that $\mathrm{Aut}(\Gamma)$ preserves the oriented labeling of the $b$labeled edges.
Consequently, there is no automorphism of $\Gamma$ which exchanges the vertices $1$ and $b$.