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. $\quad\Box$

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 $1\le k\le \lfloor n/8\rfloor=:k_n$, 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*} \begin{aligned} s_n&\ge a_*\sum_{1\le k\le k_n}(n-k)+a_{\min}\sum_{1\le k\le n-1}(n-k) \\ &=\frac{(2n-1)k_n-k_n^2}{2 \sqrt{2} \left(1-\frac{1}{\sqrt{2}}\right)^{3/2}}-\frac{n^2-n}{4 \sqrt{2}} \\ &>\frac{(2n-1)(n/8-1)-(n/8-1)^2}{2 \sqrt{2} \left(1-\frac{1}{\sqrt{2}}\right)^{3/2}}-\frac{n^2-n}{4 \sqrt{2}} \\ &>-4.006 n + 0.345 n^2 >0 \end{aligned} \end{equation*} for $n\ge12$. That $s_n>0$ for $n=5,\dots,11$ is easy to check. $\quad\Box$