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Sep 21, 2012 at 8:48 vote accept Rupert
Sep 20, 2012 at 17:22 comment added Lee Mosher One kind of regularity to impose, not as strong as biregularity, might be cocompactess. But there are counter-examples in that situation too. There exist locally finite trees where the automorphism group acts cocompactly, freely, and properly discontinuously, so all vertex stabilizers are trivial, but the automorphism group is free of finite rank.
Sep 20, 2012 at 10:40 answer added Tom De Medts timeline score: 4
Sep 20, 2012 at 9:26 comment added YCor You certainly need some regularity assumption, because the automorphism group can be trivial. Probably you want regular or biregular.
Sep 20, 2012 at 9:03 comment added Nick Gill I could be wrong but, in Section 3 of this paper - perso.uclouvain.be/pierre-emmanuel.caprace/papers_pdf/… - `Tits' independence property' is discussed. This might be what you are looking for.
Sep 20, 2012 at 8:56 history asked Rupert CC BY-SA 3.0