Does there exists a smooth transitive action of a (finite dimensional) Lie group $ G $ on the Hantzsche-Wendt manifold?
In other words, does there exists a Lie group $ G $ and a closed subgroup $ H $ such that $ G/H $ is diffeomorphic to the Hantzsche-Wendt manifold?
If such a transitive actions exists my guess is that the group $ G $ is the Euclidean group $ E_3 $ or some subgroup of $ E_3 $. Note that $ G $ must be noncompact as all three manifolds with transitive action by a compact group are already given here
https://math.stackexchange.com/a/4364430/758507
Also the group must be at least dimension 4 since all manifolds which are the quotient of a three dimensional Lie group by a cocompact lattice are given here
https://www.sciencedirect.com/science/article/pii/0166864181900183
Some background:
The Hantzsche-Wendt manifold is a compact connected flat orientable 3 manifold.
Like all compact flat manifolds it is normally covered by a torus, in this case $ T^3 $. And moreover (like all flat manifolds) it is aspherical. So it is determined by its fundamental group which is presented in https://arxiv.org/abs/math/0311476 as
$$
\pi_1(M) \cong <X,Y:X=Y^2XY^2,Y=X^2YX^2>
$$
where $ X,Y,Z=(XY)^{-1} $ are the generating screw motions which square to the translations $ t_1= X^2, t_2=Y^2,t_3=Z^2 $ given in Wolf theorem 3.5.5. Since $ M $ is compact and flat $ \pi_1 $ is a Bieberbach group, indeed it fits into the short exact sequence
$$
1 \to \mathbb{Z}^3 \to \pi_1(M) \to C_2 \times C_2 \to 1
$$
so $ M $ has holonomy $ C_2 \times C_2 $. Abelianizing $ \pi_1 $ we can see that
the first homology is
$$
H_1(M,\mathbb{Z})\cong C_4 \times C_4
$$
From the short exact sequence above we can see that $ \pi_1 $ is virtually abelian and solvable.
Best Answer
It's a little silly to post an answer right after posting the question but I thought about it more and it turns out that such an action is impossible.
The idea for arguing based on the Mostow paper is from
https://math.stackexchange.com/a/4315382/758507
Theorem C of
https://www.researchgate.net/publication/227067355_A_Structure_Theorem_for_Homogeneous_Spaces
states that a compact homogeneous space is a fiber bundle (with connected base and fiber) with base a compact Riemannian homogeneous space and fiber a compact solvmanifold unless it is the quotient of a Lie group by a cocompact lattice (what Mostow calls a HLG= homogeneous local group).
This reference classifies all compact 3d HLG
https://www.semanticscholar.org/paper/3-manifolds-whose-universal-coverings-are-Lie-Raymond-Vasquez/16a79aee94d93062d45cc03630a21b42da5d6046
So $M$ is not a HLG. So it must be a bundle of a solvmanifold over a compact Riemannian homogeneous space.
So we move on to the bundle case:
The base of this bundle cannot have dimension 3 as $ M $ is not a compact Riemannian homogeneous space (since it has no transitive action by a compact group). For example because $ \pi_1(M) $ does not have finite commutator subgroup
Transitive action by compact Lie group implies almost abelian fundamental group
So we must have a nontrivial fiber bundle:
If the fiber is dimension 3 we still have a nontrivial fiber bundle since every compact solvmanifold (except the circle) is a fiber bundle with base a torus and fiber a compact nilmanifold, both of strictly lower dimension. (fun fact: a compact nilmanifold is exactly an iterated principal circle bundle)
Since we are in such a low dimension a bundle must be either $$ S^1 \to M \to \Sigma $$ where $ \Sigma $ is Riemannian homogeneous ($ T^2,S^2,\mathbb{R}P^2 $) or the bundle must be $$ \Sigma \to M \to S^1 $$ where $ \Sigma $ is a solvmanifold (so either torus $ T^2 $ or Klein bottle $ K $). Since $ S^1 $ and $ M $ are aspherical $ \Sigma $ cannot be $ S^2 $ or $ \mathbb{R}P^2 $. So the fiber bundle must be either $$ S^1 \to M \to T^2 $$ or $$ K,T^2 \to M \to S^1 $$ But both of these are impossible because LES homotopy gives that $ \pi_1(M) $ is the semidirect product of $ \pi_1 $ of the base by normal action of $ \pi_1 $ of the fiber. Abelianizing we have that $ H_1(M) $ is direct product of $ H_1 $ of the base with a quotient of $ \pi_1 $ of the fiber. See for example
https://mathoverflow.net/questions/35713/abelianization-of-a-semidirect-product
but $ H_1 $ of the base always has free rank at least 1 for all these bundles so that would imply $ H_1(M) $ has free rank at least 1. That rules out the Hantzsche-Wendt Manifold which has first Betti number $0$.
The idea for ruling out bundles using the fact that $ M $ has vanishing first Betti number is due to
https://mathoverflow.net/a/415400/387190