Let $(M,g)$ be a non-compact homogeneous, 1-connected Riemannian manifold with $G := Isom(M,g)$ and let $v \in \Gamma (M,TM)$ be a $G$-invariant, nowhere-vashining vector field, such that $||v|| = 1$ everywhere. Then $v$ determines a global flow $\phi_v: \mathbb R \times M \to M$ such that at each time $t \in \mathbb R$, $\phi_v^t := \phi_v(t, \; \_ \;)$ commutes with every element $g \in G$
The trajectories of $\phi_v$ form the leaves of a $1$-dimensional foliation $\mathfrak F$ of $M$. Now suppose further that for each $t \in \mathbb R$, we have that $\phi_v^t \in G$. Now I have the following questions:
1) Is $\mathfrak F$ always a regular foliation? (in other words, are the trajectories embedded submanifolds ?)
2) If so, does $Y :=M/ \mathfrak F$ always have the structure of a smooth manifold, such that the quotient map is a submersion
3) For $F \in \mathfrak F$ a leaf, is $F \hookrightarrow M \rightarrow Y $ always a fiber bundle ? (In particular, are all leaves diffeomorphic? )
Consider the 3-dimensional round sphere and its isometry group $O(4)$. This group contains the subgroup $U(2)$, whose center is isomorphic to $U(1)$ (scalar unitary matrices). Of course, $O(4)\ne U(2)$, but one can modify the constant curvature metric on $S^3$ making it a Berger sphere $B=S^3_{t,\epsilon}$, whose isometry group (for generic values of $\epsilon, t$) is $U(2)$, see this paper:
P.Gadea and J.Oubina, Homogeneous Riemannian Structures on Berger 3-Spheres.
Now, take $C=B\times B$ with the product metric. Except for the $Z_2$-symmetry (swapping the factors), which one can eliminate by, say, rescaling the metric on one of the factors, the isometry group of $C$ is $U(2)\times U(2)$, hence, its center is the torus $T^2=S^1\times S^1$. The manifold $C$ is homeogeneous, simply-connected and compact. But $T^2$ contains a subgroup $H$ isomorphic to ${\mathbb R}$ (actually, continuum of such subgroups, but we just need one) whose orbits are not closed (their closures are tori); the action of this subgroup yields your flow. This is a counter-example in the setting of compact manifolds. To make a noncompact example, take $M=C\times {\mathbb R}$ with the product metric and use the same subgroup $H$ as before: The full isometry group of $M$ is $Isom({\mathbb R})\times U(2)\times U(2)$, so $H$ is still contained in its center.