We only need to consider $n\ge5$. 

Let us move each of the $n$ "equidistant points" on the equator slightly towards one of the two poles of the globe, so that after such a movement the points on the sphere become 
\begin{equation*}
	x_k(t):=\Big(\sqrt{1-t^2 z_k^2}\,\cos\frac{ 2\pi k}{n},
	\sqrt{1-t^2 z_k^2}\,\sin\frac{2\pi k}{n},t z_k\Big),
\end{equation*}
where (say) $k=0,\dots,n-1$, the $z_k$'s are real numbers (to be chosen later), and $t\ge0$ is  small. So, at $t=0$ we get the original $n$ "equidistant points" on the equator line. 

The corresponding potential energy is 
\begin{equation*}
	u(t):=\sum_{0\le i<j\le n-1}\frac1{\|x_i(t)-x_j(t)\|}, 
\end{equation*}
so that $u(0)$ is the initial potential energy, of $n$ "equidistant points" on the equator. Obviously, $u'(0)=0$. So, consider 
\begin{equation*}
	u''(0)=\sum_{0\le i<j\le n-1}
	\frac{2 z_i z_j-(z_i^2+z_j^2) \cos \dfrac{2 \pi(i-j)}{n}}
   {2^{3/2}\Big(1- \cos \dfrac{2 \pi(i-j)}{n}\Big)^{3/2}}. 
\end{equation*}

We want to choose the $z_k$'s so that $u''(0)<0$. It appears that the alternating choice $z_k:=(-1)^k$ for all $k$ will do. 

However, it is somewhat more convenient to choose the $z_k$'s at random. Namely, let the $z_k$'s be independent Rademacher random variables, so that $P(z_k=\pm1)=1/2$ for each $k$. Then, after taking the expectation, the term $z_i z_j$ disappears: 
\begin{equation*}
E	u''(0)=-\frac{s_n}{\sqrt2},
\end{equation*}
where 
\begin{equation*}
	s_n:=\sum_{0\le i<j\le n-1}a_{j-i}=\sum_{1\le k\le n-1}(n-k)a_k, 
\end{equation*}
\begin{equation*}
	a_k:=a_{n,k}:=A(k/n),\quad A(x):=\frac{\cos 2 \pi x} {(1- \cos 2 \pi x)^{3/2}}. 
\end{equation*}

So, it suffices to show that $s_n>0$ for $n\ge5$. 

Using the substitution $y:=\cos 2 \pi x$, we see that for all $x\in(0,1)$ 
\begin{equation*}
	A(x)\ge a_{\min}:=-\frac1{2\sqrt2}
\end{equation*}
Also, if an integer $k$ is such that $1\le k\le n/8$, then $\cos2\pi\frac kn\ge\cos\frac\pi4=1/\sqrt2$ and hence $a_k\ge a_*:=\dfrac{1/\sqrt2} {(1- 1/\sqrt2)^{3/2}}$. So, 
\begin{equation*}
	s_n\ge a_*\sum_{1\le k\le n/8}(n-k)+a_{\min}\sum_{1\le k\le n-1}(n-k)>0 
\end{equation*}
for $n\ge12$. That $s_n>0$ for $n=5,\dots,11$ is easy to check. $\quad\Box$