The answer is no. The explanation of Anton is of course correct but there exist stronder statements in the literature: for example by Sinjukov (Dokl. Akad. Nauk SSSR (N.S.) 98, (1954) 21--23) any symmetric space is locally \emph{geodesically rigid} is the sense that any
other metric having the same (unparameterized) geodesics with it is affinely equivalent to if (i.e., the Levi-Civita connections coincide) which in the irreducible case means that the metrics are proportional.

Actually, stronger statements hold. For example from the Lichnerowicz-Obata conjecture arXiv:math/0407337 it follows that compact Riemannian homogeneous
metrics such that sectional curvature is not constant and positive
are also geodesically rigid. Indeed, a Killing vector field for the initial metric is a infinitesimal projective transformation for the second, which must be also Killing by the projective Lichnerowicz-Obata conjecture. Then, the isometry algebras of the metrics are the same and therefore their volume forms are the same and these already implies (short tensor calculations, see for examples eqns. (1), (4), (5) of arXiv:0806.3169) that the metrics are affinely equivalent. I do not know whether homogeneous metrics of nonconstant curvature are geodesically rigid locally but all examples indicate that probably they are.

Now, in the case your metric is Kähler and not flat,

then if it is not geodesically rigid then it is locally a cone over a (sasakian) manifold which in particular implies that the manifold is not compact.
This statement is pretty nontrivial and follows from Theorem 4.6. of Mikes (Journal of Mathematical Sciences 78(1996) 311-333) combined with the Splitting Lemma from arXiv:0904.0535 and combined with the following statement which was explained to me by Kiosak and which is probably not published: Warped product Kähler nonflat
metric is a locally a cone over a sasakian manifold.

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