Here's a proof that all isotropic manifolds are homogeneous. Given any $p$ and $q$ in $M$, let $\gamma:[0,2]\rightarrow M$ be a minimizing geodesic from $p$ to $q$. Set $r = \gamma(1)$. So, following $\gamma'(1)$ along for one unit of time lands you at $q$ while following it backwards for one unit o ftime lands you at $p$.
By assumption, there is an isometry $f$ for which $d_r f$ maps $\gamma'(1) \in T_r M$ to $-\gamma'(1)$. Then, by uniqueness of geodesics, we have \begin{align*} f(q) &= f(\exp_r ( \gamma'(1)))\\ &= \exp_r(d_r f \, \gamma'(1)) \\ &= \exp_r(-\gamma'(1)) \\ &= p.\end{align*}
As other have pointed out, among, say, compact simply connected homogeneous spaces, isotropic spaces are very rare. In fact, given such a homogeneous space $G/H$ (where we assume wlog $H$ and $G$ share no common normal subgroups of positive dimension), this space is isotropic iff the induced action of $H$ on $T_{eH} G/H$ is transitive on the unit sphere. Under these assumptions, one can prove, for example, that the universal cover of $H$ has at most two factors. (In fact, those $H$ which act effectively and transitively on a sphere have been completely classified.)
I don't know of a classification of when $G/H$ has $H$ acting transitively on the unit sphere, but beyond $\mathbb{C}P^n$, it also happens for $\mathbb{H}P^n$ and $\mathbb{O}P^2$ (the compact, rank one, symmetric spaces). I don't know any other examples of homogeneous spaces for which $H$ acts transitively on the sphere.
Let $G$ be a connected Lie group and $H$ a closed subgroup (not necessarily compact).
The set $\mathcal{X}_{G,H}$ of $G$-invariant Riemannian metrics on $G/H$ is in canonical bijection with the set of $H$-invariant scalar products on $\mathfrak{g}/\mathfrak{h}$ under the adjoint representation of $G$. If $H$ is compact, the latter is non-empty.
Examples:
- If $H$ is trivial, $\mathcal{X}_{G,H}$ is thus quite big (a non-empty open cone in dimension $n(n+1)/2$ where $n=\dim(G)$). More generally, if $H$ is normal, this still holds (with $n=\dim(G/H)$).
- But there are cases where $\dim(G/H)>1$ and it's a half-line, i.e., the $G$-invariant Riemannian metric on $G/H$ is unique up to rescaling. This holds, in particular, when $G$ is a simple Lie group and $H$ is a maximal compact subgroup. It also holds when $G$ is the isometry group of some irreducible symmetric space of compact type, which includes the case of $\mathrm{SO}(3)/\mathrm{SO}(2)$.
- If $H$ is compact, then $\mathcal{X}_{G,H}$ is not empty.
- If $H$ is non-compact, then $\mathcal{X}_{G,H}$ can be non-empty (as when $H$ is normal, see above), or can be empty, for instance when $\dim(G)=2$ and $H$ is a non-normal 1-parameter subgroup.
- If $G=\mathrm{SL}_2(\mathbf{R})$ (or more generally any simple Lie group) and $H$ is an arbitrary non-compact closed subgroup, then $\mathcal{X}_{G,H}$ is empty ($\star$).
($\star$): assume that $H$ is non-compact (so $G$ is non-compact) and that $G/H$ carries an invariant Riemannian metric. Since $H$ is non-compact, the action of $G$ on $G/H$ is non-proper. Hence it has bounded orbits (reference), so $G/H$ is compact. Since moreover it is Riemannian, it has finite volume and it follows that $H$ is a cocompact lattice; in particular it is Zariski-dense. Since the stabilizer of a Riemannian metric of $\mathfrak{g}/\mathfrak{h}=\mathfrak{g}$ is Zariski-closed, it follows that the metric on $\mathfrak{g}$ is $G$-invariant for the adjoint representation. Since $G$ is non-compact, there is no such metric (the set of invariant quadratic forms being 1-dimensional, generated by the Killing form) and we get a contradiction.
Best Answer
A Riemannian homogeneous manifold is diffeomorphic to the product of a Euclidean space with a compact Riemannian homogeneous manifold. See
https://mathoverflow.net/questions/410334/noncompact-riemannian-homogeneous-is-trivial-vector-bundle-over-compact-homogene
So it is enough to show this for the compact case.
Let $ M $ be a flat compact Riemannian homogeneous manifold. Then the fundamental group of $ M $ is a torsion free crystallographic group (also known as a Bieberbach group). Since $ M $ is compact and Riemannian homogeneous then by
Transitive action by compact Lie group implies almost abelian fundamental group
the commutator subgroup of the fundamental group is finite. But the fundamental group is torsion free so every finite subgroup is trivial. Thus the commutator subgroup must be trivial. In other words $ \pi_1(M) $ is abelian. So it must be $ \mathbb{Z}^n $. So $ M $ must be isometric to a flat torus.