It is a known fact (and it has been discussed several times here on MSE) that if $p, q$ are points of $\mathbf{R}^n$ and $\sigma$ is a (piecewise $\mathcal{C}^1$, rectifiable) curve connecting them, then $$l(\sigma)\ge |q-p|$$ and if $\sigma$ is a segment of a straight line equality holds. But is the converse true? If $\sigma$ is a (piecewise $\mathcal{C}^1$) curve connecting $p$ and $q$ satisfying $$l(\sigma)=|q-p|,$$ is it true that $\sigma$ is (up to diffeomorphism) a straight line?
Please notice that I am referring only to $\mathbf{R}^n$ with its standard (Euclidean) norm, no Riemannian metrics or submanifolds involved.
If $\sigma\colon [0,1]\to \Bbb R^n$ is your curve, then $l(\sigma)$ is defined as supremum of all $|\sigma(t_1)-\sigma(t_0)|+|\sigma(t_2)-\sigma(t_1)|+\cdots +|\sigma(t_n)-\sigma(t_{n-1})|$ over all tuples $(t_0,\ldots, t_n)$ with $0= t_0<t_1\ldots <t_n=1$.
Now suppose that for some $t^*\in[0,1]$, $\sigma(t^*)$ is not on $pq$. Then $$l(\sigma)\ge |\sigma(t^*)-p|+|q-\sigma(t^*)|>|p-q|.$$