A property of bi-invarient Finsler metrics on SU(N) Does the set of bi-invariant Finsler metrics on $SU(N)$ exactly coincide the set of Finsler metrics with the one-parameter subgroups as their geodesics through the identity?
I know that being bi-invariant implies that the geodesics through the identity are exactly the one parameter subgroups. That is to say, every element of the Lie algebra is a geodesic vector. However, I can't seem to settle whether or not the converse holds.
 A: 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 the paper Periodic solutions of Hilbert's fourth problem.
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 bit 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 (https://arxiv.org/abs/math/0601522), this is known to imply that $(G,F)$ is a symmetric space and so that the metric $F$ is bi-invariant.
A: Yes, it is. If  affinely parameterized geodesics    of two finsler metrics are the same, and one of them is Berwald, the other is Berwald as well. This fact is known, 
 I believe the explanation is on   p. 74 of  [S.S. Chern and Z. Shen: Riemann-Finsler geometry. Nankai Tracts in Mathematics 6. World Scientific (2005)], at least this is written in some paper of mine  but now I do not have the  book by hand so I can not check it. But it is true.  
In your case, each geodesic of the finsler metric is a geodesic of the  bi-invariant Riemannian metric which is Berwald and therefore  the initial metric is Berwald as well. Its holonomy group is the same as the holonomy group of the  standard bi-invariant Riemannian metric and since it acts transitively on the tangent space the finsler  metric is Riemannian. 
A: It depends if by "geodesics" you mean parameterized geodesics or unparameterized geodesics. If you mean "parameterized", then Matveev's answer will do. If you mean unparameterized geodesics, then there are more metrics whose geodesics agree, as point sets, with one-parameter subgroups. This is easily seen in the case of $SU(2)$, which is diffeomorphic to the $3$-sphere, and there are lots of metrics whose unparameterized geodesics are great circles (Hilbert's fourth problem). 
