Suppose there is a trajectory through a 2D Euclidean plane:
$$\begin{eqnarray} r :\ &\mathbb{R} &\to \mathbb{R}^{2} \\ &t &\mapsto \left(x(t),y(t)\right) \end{eqnarray}$$
The length of this trajectory is the functional
$$ l[r] = \int\mathrm{d}t\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^{2} + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^{2}}$$
When you vary $l[r]$ keeping the endpoints fixed, then when $\frac{\delta l}{\delta r} = 0$ $r$ is a straight line between the endpoints. For any non-extremal length (i.e., any non-straight trajectory) between the endpoints, there are always at least two possible trajectories with that length in $\mathbb{R}^{2}$. In $\mathbb{R}^{3}$ there are an infinite number of non-straight trajectories with any given length.
Now, in physics we often use the similar functional
$$ A[r] = \int\mathrm{d}t\left[\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^{2} + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^{2}\right] $$
(with some mass coefficient that's unimportant when dealing with a single particle) to find the trajectory of a free particle. Obviously $A[r]$ has the same extremum as $l[r]$ because the positive square root is a monotonic positive function, and this is how the action is often motivated. However, I'm interested in the non-extremal paths, specifically: does using the action eliminate some of the redundancy that is present in the length?
Or, to put it another way: if there are two paths with the same length and the same endpoints, $l[r] = l[r']$, will their actions be equal, $A[r] \overset{?}{=} A[r']$?
Apologies for any mathematical imprecision in the question.