A square on the equator of a sphere is a critical point of the electrostatic potential $\newcommand{\S}{\mathbb{S}^2}$
This is a self-answered question. I learned something from spelling out the details, and I hope this could be interesting to others. I would welcome alternative solutions.
Let$$M=\{(x_1,x_2,x_3,x_4) \in  \mathbb{S}^2 \times \mathbb{S}^2 \times \mathbb{S}^2 \times \mathbb{S}^2 \, |\,\, \text{ all the } x_i \, \text{ are distinct}\} $$
Let $E:M \to \mathbb{R}$ be defined by
$$E(x_1,x_2,x_3,x_4)=\sum_{i < j}\frac{1}{\| x_i - x_j \|},$$
where $\| x_i - x_j \|$ denotes the Euclidean distance in $\mathbb{R}^3$.
Claim: Let $p:=(p_1,p_2,p_3,p_4)$ be the configuration of a planar square lying on the equator of $\S$. Then $p$ is a critical point of $E$.
Question:
Is there an "easy" proof for that claim, that does not rely on explicit computation of the differential $dE$?
 A: $\newcommand{\S}{\mathbb{S}^2}$
We give a symmetry-based argument due to Kajelad. The proof only uses the property that $E$ is invariant under isometries.
Let $R:\mathbb{R}^3 \to \mathbb{R}^3$ be the reflection across the equatorial plane $z=0$, i.e. $R(x,y,z)=(x,y,-z)$.
$R$ induces a map $\tilde R:(\mathbb{S}^2)^4 \to (\mathbb{S}^2)^4$, given by
$$
\tilde R(x_1,x_2,x_3,x_4):=\big(R(x_1),R(x_2),R(x_3),R(x_4)\big).
$$
Since $R,\tilde R$ are isometries, we have $E \circ \tilde R = E$, and thus
$$
dE_{p}=dE_{\tilde R(p)} \circ d\tilde R_p =dE_{\tilde R(p)} \circ \tilde R .
$$
Set $e_3=(0,0,1) \in T_{p_1}\S$. Then using $\tilde R(p)=p$ ($R(p_i)=p_i$) we get
$$
dE_{p}(e_3,0,0,0)=dE_{\tilde R(p)} \circ \tilde R(e_3,0,0,0)=dE_{p}(-e_3,0,0,0),
$$
so
$$
dE_{p}(e_3,0,0,0)=0. \tag{1}
$$

We turn to the second part of the argument:
The idea in a nutshell:
We have two symmetries which fixes $p_1$: one a diagonal reflection/swap $R$, and one "general coordinate-exchange" $P$ ($x_2 \iff x_4$). Since $dR$ acts as minus map on $T_{p_1}\S$ and $dP$ acts as the identity, the relevant derivative should be zero.
Here are the details:
Consider the diagonal reflection $R$ which swaps $p_2$ and $p_4$. (we can extend $R$ to be the identity on $e_3=(0,0,1)$, or restrict the discussion to the plane of the equator). Then
$$
dE_{p}=dE_{\tilde R(p)} \circ \tilde R .
$$
Let $v \in T_{p_1}\S$ be tangent to the equator. Then
$$
dE_{p}(v,0,0,0)=dE_{\tilde R(p)}(-v,0,0,0). \tag{2}
$$
On the other hand, let $P(x_1,x_2,x_3,x_4)=(x_1,x_4,x_3,x_2)$. Since $P$ is an isometry, we have
$$
E \circ P=E, 
$$
so
$$
dE_{p}(v_1,v_2,v_3,v_4)=dE_{P(p)} \circ P(v_1,v_2,v_3,v_4)=dE_{P(p)}(v_1,v_4,v_3,v_2).
$$
In particular,
$$
dE_{p}(v,0,0,0)=dE_{P(p)}(v,0,0,0). \tag{3}
$$
Writing explicitly equations $(2)-(3)$, using $p=(p_1,p_2,p_3,p_4)$, we get
$$
dE_{(p_1,p_2,p_3,p_4)}(v,0,0,0)=dE_{(p_1,p_4,p_3,p_2)}(-v,0,0,0), \tag{2'}
$$
$$
dE_{(p_1,p_2,p_3,p_4)}(v,0,0,0)=dE_{(p_1,p_4,p_3,p_2)}(v,0,0,0).  \tag{3'}
$$
Combining the last two equations results in
$$
dE_{(p_1,p_2,p_3,p_4)}(v,0,0,0)=0. \tag{4}$$
Equations $(1),(4)$ imply that
$$
dE_{p}(v,0,0,0)=dE_{p}(e_3,0,0,0),
$$
where $v \in T_{p_1}\S$ is tangent to the equator. Since $\text{span}\{e_3,v\}=T_{p_1}\S$, it follows that
$$
dE_{p}(w,0,0,0)=0
$$
for any $w \in T_{p_1}\S$. By symmetry, this is true for the other three "points/components", so $dE_p=0$ as required.
