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Feb 24, 2012 at 11:10 comment added BS. @Paul : You should perhaps say "no compact connected Lie group", as there might be finite order symmetries. This is equivalent to "no non trivial $S^1$ action".
Feb 24, 2012 at 3:13 comment added Vitali Kapovitch there are indeed many such 4-manifolds. connected sum of any number of $K3$-surfaces is the most standard example. so if anything like this is a homogenous space $G/H$ then the maximal compact subgroup of $G$ must be zero-dimensional which means that $G$ must be contractible. This is of course impossible which means that none of such manifolds can be homogeneous.
Feb 24, 2012 at 3:03 comment added Paul Siegel I guess my answer doesn't rule out the possibility that $M$ is a homogeneous space for a non-compact Lie group, but perhaps it's still interesting.
Feb 24, 2012 at 3:02 history answered Paul Siegel CC BY-SA 3.0