Classify cohomogeneity one actions up to what? This is a question about cohomogeneity one actions of a compact Lie group $G$ on a Riemannian manifold $M$, such that $G$ acts via isometries.
Many articles have been published about classifications of such actions.
Different articles classify actions up to different notions of equivalence.
Here are two examples:


*

*Classification up to $G$-diffeomorphism (i.e. there exists a diffeomorphism $M \rightarrow M$ that is $G$ equivariant, i.e. $f(g \cdot_1 x)=g \cdot_2 f(x)$, where $\cdot_1$ and $\cdot_2$ denote two different actions of $G$ on $M$):
Karsten Grove, Burkhard Wilking & Wolfgang Ziller: POSITIVELY CURVED COHOMOGENEITY ONE MANIFOLDS AND 3-SASAKIAN GEOMETRY

*Classification up to isometric orbit equivalence (i.e. there exists an isometry $f:M \rightarrow M$ such that $G \cdot_1 f(x)=f(G \cdot_2 x)$):
J. Berndt M. Domínguez-Vázquez: COHOMOGENEITY ONE ACTIONS ON SOME NONCOMPACT SYMMETRIC SPACES OF RANK TWO

*The notion of "$G$-isometry" (replace the diffeomorphism in point 1 by an isometry) would be the most natural to me, but I haven't found it in the literature.

Question 1:
  Are these notions identical?
Question 2:
  One application of a classification of cohomogeneity one actions was to find an exotic nearly Kähler structure on $S^6$ (Lorenzo Foscolo, Mark Haskins: New $G_2$-holonomy cones and exotic nearly Kähler structures on $S^6$).
  They used a classification of cohomogeneity one actions on $S^6$ up to $G$-diffeomorphism.
  Could it happen that there is another, hidden exotic nearly Kähler structure on $S^6$ which arises from a $G$-diffeomorphic cohomogeneity one action on $S^6$ that is not $G$-isometric?
  (That would lead to two nearly Kähler structures that are not isometric)

 A: First, for the isometry vs diffeomorphism issue, diffeomorphism is more natural, in some sense.   For example, if one considers the standard $S^1$ action on $S^2$ by rotations, there are many metrics on $S^2$ for which this action is isometric - if one deforms $S^2$ into a football shape, the $S^1$ action is isometric.
More generally, any way of getting something diffeomorphic to $S^2$ as a surface of revolution admits an isometric cohomogeneity one action by $S^1$.
On the other hand, all these seemingly different pictures are equivariantly diffeomorphic.
This same idea works on any cohom 1 manifold $M$ which is compact.  Specifically, the quotient space $M/G$ is either $S^1$ or $[0,1]$.  In the first case, $M$ looks like a $G/H$ bundle over $S^1$ for some homogeneous space $G/H$.  One can scale the metric on the $G/H$ fibers by different amounts as one traverses the circle, so there are always many $G$-invariant metrics.  In the second case, $M$ is a union of two disc bundles with boundary a homogeoneous space $G/H$.  That is, $M = G/H\times [0,1]$ with some quotienting happening at both end points.  Then one must be careful with the metric near the boundary (to keep everything smooth), but can freely scale the metric on $G/H$.
(In addition, often $G/H$ has many families of deformations of metrics with are all $G$-invariant.  This would have to be dealt with in any classification up to $G$-isometry).
Now, for question 1, no, these are not the same.  For a really dumb example, suppose $G$ acts on $M$ via a cohomogeneity 1 action: $g\ast m = gm$.  Consider the two different $G\times G$ actions on $M$:  $(g_1,g_2)\ast_1 m = g_1 m$ and $(g_1,g_2)\ast_2 m = g_2 m$.  Trivially, these actions are orbit equivalent, but they are not $G\times G$-equivariant.
To see that they are not $G$-equivariant, let $f$ be a $G$-equivariant diffeomorphism.  Choose $m\in M$ for which $f(m)$ is not fixed by  $G$, and let $e\in G$ denote the identity.  Then for any $g_2\in G$,  $f(m) = f(m) = f((e,g_2) \ast_1 m) = (e,g_2)\ast_2 f(m)$, so $f(m)$ is fixed by $G$.
For a slightly less dumb idea, consider the $S^1$ action on $S^2$ via rotations compared to the $S^1$ action on $S^2$ via rotations going twice as fast.  Then these are clearly orbit equivalent (orbits are lines of latitude) but they are not $G$-equivariant since, for example, the kernels of the actions are different.
(I do not know of a non-"dumb" idea in the cohom 1 case which distinguishes the notions of $G$-equivariant from orbit-equivalent.)
For question 2, I don't know.
