I am re-editing this response because I got it wrong the first time around.
The following result, due to myself and José Barbosa Gomes, is (a small) part of a paper that will appear in ArXiv on Tuesday (9/11/2018) or thereabouts.
Theorem. If a C2 reversible Finsler metric $F$ on a compact, connected Lie group which is neither $SU(2)$ nor $SO(3)$ has the same unparametrized geodesics as a bi-invariant Riemannian metric, then it is a bi-invariant metric.
So the answer to your question is yes for $N > 2$. Let me recall that a reversible Finsler metric is one for which the length of each tangent vector $v_x$ equals the length of its opposite $-v_x$.
If you admit the metric to be smooth (= as smooth as needed with the smoothness depending on the dimension) I can give you a sketchy shortcut to the proof in the paper.
First one proves that a smooth reversible Finsler metric whose geodesics are straight lines on a torus of dimension $k$ $(k > 1)$ is flat (= locally isometric to a normed space). This is done by looking a but under the hood at the Busemann-Pogorelov solution of Hilbert's fourth problem.
Now if $G$ is neither $SU(2)$ nor $SO(3)$, then it has rank greater than one as a symmetric space when provided with a bi-invariant Riemannian metric. This means every geodesic belongs to a flat torus of dimension at least two. If you have a Finsler metric $F$ with the same unparametrized geodesics, then those tori will be totally geodesic for $F$ and, by the result above, the restriction of F to those tori will also be a flat metric. From this we deduce that $F$ is not just projectively equivalent to the bi-invariant Riemannian metric, but also affinely equivalent to it (i.e., the midpoints of geodesic segments are the same). By work of Z.I. Szabo (http://front.math.ucdavis.edu/0601.5522), this is known to imply that $(G,F)$ is a symmetric space and so that the metric $F$ is bi-invariant.