Now that I have a better understanding of what the question entails, I am going reorganize my answer. We are interested in points where a function of form $v(\mathbf{x}):=u(\mathbf{x})-u(R\mathbf{x})$--here $R\in{\rm{SO}}(3)$ denotes a rotation--and its gradient vanish simultaneously; i.e. points $\mathbf{x}_0\in\Bbb{S}^2$ with $u(\mathbf{x}_0)=u(R\mathbf{x}_0)$ and
$\nabla u(\mathbf{x}_0)-R^{\rm{T}}\nabla u(R\mathbf{x}_0)=\mathbf{0}\Leftrightarrow
R\nabla u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$. Of course, certain trivial cases should be excluded, like $R$ being the identity or more generally, cases where $R\mathbf{x}_0=\mathbf{x}_0$ is allowed: Otherwise, one can take $\mathbf{x}_0$ to be a critical point and any rotation fixing the unit vector $\mathbf{x}_0$ works. Hence I am going to focus on cases where $R\mathbf{x}_0\neq\mathbf{x}_0$. The question is if there exists $R\in{\rm{SO}}(3)$ and **two** distinct points on the sphere for which these equations hold.

**Question)** Are there two distinct points $\mathbf{x}_0,\mathbf{x}_1\in\Bbb{S}^2$ and a rotation $R\in{\rm{SO}(3)}$ such that $R\mathbf{x}_i\neq\mathbf{x}_i$, $u(\mathbf{x}_i)=u(R\mathbf{x}_i)$ and
$R\nabla u(\mathbf{x}_i)=\nabla u(R\mathbf{x}_i)$ for $i\in\{0,1\}$?

*I am going to prove the existence of one such point, and discuss how the same idea can potentially be applied to the general question.* But before that, a remark regarding the special case of $180^\circ$-rotations $R=R_{\mathbf{y}}$ which are mentioned at the beginning of the question: If $R=R_{\mathbf{y}}$, since this is an involution, once we find one desired point $\mathbf{x}_0$, $R_{\mathbf{y}}\mathbf{x}_0$ would work as $\mathbf{x}_1$. Notice that $R_{\mathbf{y}}\mathbf{x}_0\neq\mathbf{x}_0$ amounts to $\mathbf{y}\neq\pm\mathbf{x}_0$ (such a condition also appears at the beginning of the question). An even more special case is when $R(\mathbf{x})=-\mathbf{x}$. I am going to say a few words about the antipodal case at the end.

**Claim)** There exist $R\in{\rm{SO}}(3)$ and $\mathbf{x}_0\in\Bbb{S}^2$ such that $R\mathbf{x}_0\neq\mathbf{x}_0$,
$u(\mathbf{x}_0)=u(R\mathbf{x}_0)$ and $R\nabla
u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$.

**Proof of the Claim)** Let $C$ be a connected component of a regular fiber of $u:\Bbb{S}^2\rightarrow\Bbb{R}$. Thus $C$ is homeomorphic to a circle. We deduce that the function $|\nabla u|:C\rightarrow (0,\infty)$ cannot be injective. So there exist points $\mathbf{x}_0\neq\mathbf{x}_1$ on $C$ with $|\nabla u(\mathbf{x}_0)|=|\nabla u(\mathbf{x}_1)|>0$. And of course $u(\mathbf{x}_0)=u(\mathbf{x}_1)$ since these points belong to the same level set. Now, given that we are dealing with the gradient of a function defined on the unit sphere, we have the following pairs of orthogonal unit vectors in $\Bbb{R}^3$:
$$
\mathbf{x}_0\perp \frac{\nabla u(\mathbf{x}_0)}{|\nabla u(\mathbf{x}_0)|}, \quad \mathbf{x}_1\perp \frac{\nabla u(\mathbf{x}_1)}{|\nabla u(\mathbf{x}_1)|}.
$$
Hence there exists a unique element $R$ of ${\rm{SO}}(3)$ with $R(\mathbf{x}_0)=\mathbf{x}_1$ and
$R\left(\frac{\nabla u(\mathbf{x}_0)}{|\nabla u(\mathbf{x}_0)|}\right)=\frac{\nabla u(\mathbf{x}_1)}{|\nabla u(\mathbf{x}_1)|}$. The latter equality can be written as $R\nabla u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$ because
$|\nabla u(\mathbf{x}_0)|=|\nabla u(\mathbf{x}_1)|$.

The partial solution above allows us to reformulate the question, at least for regular points of $u$, as:

**Question$^\prime$)** Let $$ \mathbf{M}:=\{(\mathbf{x},\mathbf{y})\in\Bbb{S}^2\times\Bbb{S}^2\mid\mathbf{x}\neq\mathbf{y},u(\mathbf{x})=u(\mathbf{y}),|\nabla
u(\mathbf{x})|=|\nabla u(\mathbf{y})|>0\}; $$ and define
$F:\mathbf{M}\rightarrow{\rm{SO}}(3)$ by setting
$F(\mathbf{x},\mathbf{y})$ to be the unique element of ${\rm{SO}}(3)$
under which $\mathbf{x}\mapsto\mathbf{y}$, $\nabla
u(\mathbf{x})\mapsto\nabla u(\mathbf{y})$ and $\mathbf{x}\times\nabla
u(\mathbf{x})\mapsto\mathbf{y}\times\nabla u(\mathbf{y})$. Is $F$
non-injective?

(Keep in mind that if two distinct pairs $(\mathbf{x}_0,\mathbf{y}_0)$ and $(\mathbf{x}_1,\mathbf{y}_1)$ belong to the same fiber $F^{-1}(R)$, then, as desired, $R\mathbf{x}_i\neq\mathbf{x}_i$, $u(\mathbf{x}_i)=u(R\mathbf{x}_i)$ and
$R\nabla u(\mathbf{x}_i)=\nabla u(R\mathbf{x}_i)$ for $i\in\{0,1\}$ where $\mathbf{x}_0\neq\mathbf{x}_1$. Also notice that $F(\mathbf{y},\mathbf{x})=F(\mathbf{x},\mathbf{y})^{-1}$, thus it is also enough to show that the range of $F$ contains an involution.)

Given that the subset $\mathbf{M}$ of $\Bbb{S}^2\times\Bbb{S}^2$ is defined by two constraints, it should generically be a 2D "object" (except in special cases like $u$ being constant). So I struggle to see any immediate argument showing that the map $F$ from $\mathbf{M}$ to the 3D group ${\rm{SO}}(3)$ cannot be injective.

**The antipodal case)** It is possible to find antipodal points belonging to same level set whose gradient vectors have the same projection along a given direction.

To see this, it suffices to apply the Borsuk–Ulam to
$$
\mathbf{x}\in \Bbb{S}^2\mapsto \left(u(\mathbf{x}),\langle\nabla u(\mathbf{x}),\mathbf{p}\rangle\right)\in\Bbb{R}^2.
$$
where $\mathbf{p}\in\Bbb{S}^2\subset\Bbb{R}^3$ is an arbitrary unit vector. The theorem implies the existence of $\mathbf{x}_0\in\Bbb{S}^2$ for which $u(\mathbf{x}_0)=u(-\mathbf{x}_0)$ and moreover,
the projections of
$\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)\in\Bbb{R}^3$
along the line parallel to $\mathbf{p}$ passing through the origin coincide.

A bit more can be said in view of the fact that $\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)$ both belong to the subspace
${\rm{T}}_{\mathbf{x}_0}\,\Bbb{S}^2={\rm{T}}_{-\mathbf{x}_0}\,\Bbb{S}^2$
which is the orthogonal complement of $\mathbf{x_0}$ in $\Bbb{R}^3$. For instance, if $\mathbf{p}$ is so that $u(\mathbf{p})\neq u(-\mathbf{p})$, then the unit vectors $\mathbf{x}_0,\mathbf{p},-\mathbf{p}$ are distinct. Hence vectors $\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)$ are subjected to two independent constraints: they are orthogonal to $\mathbf{x}_0$ and have the same projection along $\mathbf{p}$; thus they differ by a multiple of $\mathbf{p}\times\mathbf{x_0}$.

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