Because you're assuming the group is compact, there's another method for classifying such spaces. This approach can be used in higher dimensions (though people often assume the resulting manifold is simply connected just to cut down on the work).
I'll also always be assuming things are connected: the components of a homogeneous space $G/H$ are always diffeomorphic, and the identity component of $G$ acts transitively on a connected component of $G/H$, so this is all ok.
Suppose $G$ is a connected compact Lie group, $H\subseteq G$ is a closed subgroup. Because $G$ is compact, we may equip the homogeneous space $G/H$ with a Riemannian metric for which the action by $G$ is isometric.
Now, consider the isotropy action of $H$ on $T_{eG} G/H$, where $e\in G$ denotes the identity element. Recall (see, e.g., this MSE question) that if an isometry of a (connected) Riemannian manifold fixes a point and acts as the identity at the tangent space at that point, then that isometry must be the identity. Elements of $h$ fix $e H\in G/H$, so, if such an $h$ acts trivially on $T_{eG}G/H$, we must have $h = e$.
Said another way, the isotropy action gives an injective map $H\rightarrow O(T_{eG} G/H)$ into the orthogonal group. Thus, we may view $H$ as a subgroup of $O(T_{eG} G/H)$.
I'll now specialize to the case where $G/H$ is $3$-dimensional, so $O(T_{eG} G/H)$ can be identified with $O(3)$. The connected subgroups of $SO(3)$ are well known: they are $\{e\}, SO(3)$, or a conjugate of the usual $SO(2)\subseteq SO(3)$. Note that $G/H$ is equivariantly diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$, so, in terms of classification, we can assume the identity component $H^0$ of $H$ is given by one of $\{e\}, SO(2)$, or $SO(3)$.
We'll break into cases depending on $H^0$.
Case A
If $H^0 = \{e\}$, then $3 = \dim G - \dim H$ impies that $\dim G = 3$. From the classification of simply connected Lie groups, $G$ has a cover which is either isomorphic to $SU(2)$ or to $T^3$. In the first case $G = SU(2)$ or $SO(3)$ and $SO(3)/H = SU(2)/\pi^{-1}(H)$ with $\pi:SU(2)\rightarrow SO(3)$ the double cover, so you just get quotients of $SU(2)$. In the second case, you get quotients of $T^3$. However, $H\subseteq T^3$ is normal, so $T^3/H$ is an abelian Lie group of dimension $3$, so $T^3/H\cong T^3$ no matter what $H$ is (so long as $H^0 = \{e\}$.)
Case B
If $H^0 = SO(2)$, then $\dim G = 4$. From the classification of simply connected Lie groups, $G$ has a cover of the form $G = T^4$ or $G = SU(2)\times S^1$. If $G = T^4$, the argument from the previous paragraph establishes that $T^4/H\cong T^3$. So, we will assume that $G$ is covered by $G = SU(2)\times S^1$. In fact, we'll pull everything back: $G/H\cong (SU(2)\times S^1)/\pi^{-1}(H)$, so we'll actually work with $G= SU(2)times S^1$. (Note though, that by pulling back, $\pi^{-1}(H)$ need not act effectively on $(SU(2)\times S^1)/\pi^{-1}(H)$) any more.) Also, instead of writing $\pi^{-1}(H)$ everywhere, I'll abuse notation and just write $H$.
Lemma: Suppose $H$ acts diagonally on $G_1\times G_2$ and that the action of $H$ on $G_2$ is transitive with isotropy group $H'$. Then $(G_1\times G_2)/H$ is canonically diffeomorphic to $G_1/H'$.
Proof: Just check that the map $G_1/H'\rightarrow (G_1\times G_2)/H$ sending $g_1 H'$ to $(g_1,e)H$ is a diffeomorphism. $\square$
Using the lemma, it follows that if the projection of $H^0$ to the $S^1$ factor of $G$ is surjective, then we can write $G/H\cong SU(2)/H'$, where $H'$ is $0$-dimensional. These were classified in Case A). So, we may assume that the projection of $H^0$ to the $S^1$ factor of $G$ is trivial. That is, we may assume $H^0$ acts only on the $SU(2)$ factor of $G$, so $G/H^0\cong S^2\times S^1$.
So, we understand $H^0$ in this case, but what about $H$? Well, let's assume $H\neq H^0$, and pick $(h_1,h_2)\in H\setminus H_0$. Because the normalizer $N:=N_{SU(2)}(SO(2))$ has two components, and $h_1 \in N$, we know that $(h_1,h_2)^2$ acts as a rotation on just the $S^1$ factor of $G/H^0 \cong S^2\times S^1$. Hence, $(G/H^0)/\langle (h_1,h_2)^2\rangle \cong S^2\times S^1$, so we may as well assume that $(h_1,h_2)^2 \in H^0$, so that, in particular, $h_2 = \pm 1$. That is, we may as well assume that $H$ has precisely two components. If $h_1\in SO(2)$, then $G/H$ is diffeomorphic to $S^2\times S^1$ independent of $h_2$. If $h_1\notin SO(2)$ and $h_2 = 1$, we get $\mathbb{R}P^2\times S^1$, and if $h_1\notin SO(2)$ and $h_2 = -1$, we get the non-trivial $S^2$-bundle over $S^1$ (the quotient of $S^2\times S^1$ by the diagonal antipodal action).
This concludes case B.
Case C In this case, $H^0 = SO(3)$. Then $\dim G = 6$, so $G$ is covered by one of $T^6, SU(2)\times T^3$, or $SU(2)\times SU(2)$.
For $G=T^6$, because $H^0$ is non-abelian, there is no $H^0\subseteq G$, even up to cover.
For the other two, by replacing $H^0$ with a cover of $H^0$ (i.e., $SU(2)$), we can actually assume $G= SU(2)\times T^3$ or $G = SU(2)\times SU(2)$. However, in both cases, it is easy to see that the above Lemma applied, so we find $G/H\cong T^3/H'$ or $SU(2)/H'$ where $H'$ is $0$-dimensional, getting us back to Case A. This completes Case C.
$\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Reals}{\Number{R}}\newcommand{\Cpx}{\Number{C}}\newcommand{\Proj}{\mathbf{P}}\newcommand{\CSum}{\mathop{\#}}$All four manifolds are double-covered by the product $S^{2} \times S^{1}$. These coverings may be represented conveniently by writing $(x, z)$ for the general element of $S^{2} \times S^{1} \subset \Reals^{3} \times \Cpx$:
- The quotient by $(x, z) \mapsto (-x, z)$ is $\Reals\Proj^{2} \times S^{1}$. A fundamental domain is a closed hemisphere times the circle.
- The quotient by $(x, z) \mapsto (-x, -z)$ is the mapping cylinder of the antipodal map of $S^{2}$. A fundamental domain is the sphere cross half the circle, i.e., $S^{2} \times [0, 1]$, and $(x, 0) \sim (-x, 1)$.
- The quotient by $(x, z) \mapsto (-x, \bar{z})$ is the connected sum $\Reals\Proj^{3} \CSum\Reals\Proj^{3}$. A fundamental domain is the sphere cross half the circle, this time with boundary identification $(x, 0) \sim (-x, 0)$ and $(x, 1) \sim (-x, 1)$. To see this is a connected sum of two $\Reals\Proj^{3}$s, note that projective space itself may be viewed as a closed ball in $\Reals^{3}$ with antipodal identification on the boundary. Removing a ball from this amounts to removing a concentric ball, i.e., identifying $(x, 0) \sim (-x, 0)$ in $S^{2} \times [0, 1]$.
- Just for completeness, $(x, z) \mapsto (x, z^{2})$ is a double-covering of $S^{2} \times S^{1}$ over itself.
Incidentally, the isometry group of a flat Klein bottle does not act transitively: A flat Klein bottle may be viewed as a quotient of the flat cylinder $\Reals \times S^{1}$ under the mapping $(t, z) \mapsto (t + 1, \bar{z})$. Consequently, an isometry of the Klein bottle lifts to an invariant isometry of the cylinder, and conversely every invariant isometry of the cylinder descends. Translations along the $\Reals$ factor descend, as does the reflection $(t, z) \mapsto (t, \bar{z})$, but rotations of the $S^{1}$ factor do not. Geometrically, there are two distinguished circles on a flat Klein bottle coming from central circles on two flat Möbius strips. The "nearby" local sections of the non-trivial circle bundle over $S^{1}$ are topologically circles mapping $2$-to$1$ to the "central" circles.
The isometries of $\Reals\Proj^{3} \CSum\Reals\Proj^{3}$ may be analyzed similarly; here we seek isometries of $\Reals \times S^{2}$ commuting with $(t, x) \mapsto (t + 1, -x)$. Here, every orthogonal transformation of Euclidean three-space descends to an isometry of the quotient.
Best Answer
A sufficient condition is that the isometry $\varphi$ is central in $G$, since in that case the action of $G\times\mathbb{R}$ preserves fibers of the covering. This is what happens for the mapping tori of the antipodal maps on round $S^n$.