Could a Riemannian metric be uniquely determined by its exponential map? It is known that isometric metrics have the same exponential maps. I am interesting in the converse, can we recover the metric from the exponential map?
Suppose that $(M,g_1)$ and $(M,g_2)$ are Riemannian manifolds with the same exponential maps, i.e., the maps $\exp_{1}:TM\rightarrow M$ and $\exp_{1}:TM\rightarrow M$ are equal. Does it follow that $(M,g_1)$ is isometric to $(M,g_2)$? 
If someone have already asked this question I will appreciate the answer. 
 A: This is a good and classical question, going back to Levi-Civita himself. I will say that two Riemannian metrics with the same Levi-Civita connection form an LC-pair. 
The first thing to observe is that two Riemannian metrics on the given manifold have the same exponential map iff they forms an LC pair. One direction is easy, as geodesics on a Riemannian manifold are curves defined by the differential equation $\nabla_{c'} c'=0$ which clearly depends only on the affine connection $\nabla$. The opposite direction is harder and I will not need it. It is proven (for instance) in: 
M.Spivak, "Comprehensive Introduction to Differential Geometry" (Publish Or Perish, 2000), volume 2, chapter 6, Appendix 1.
Thus, your question can be reformulated as: 
Suppose that $g_1, g_2$ are Riemannian metrics on a manifold $M$ forming an LC pair. 
Does it follow that $(M,g_1)$ is isometric to $(M,g_2)$?
As one easily observes (see the comment by @Harald Hanche-Olsen), if $g_1$ is a constant multiple of $g_2$ (with a factor $a\ne 1$), they form an LC pair (I will refer to such LC pairs as trivial) but, typically, are not isometric. For instance, if $M$ is compact then $(M,g_1), (M,g_2)$ have different volumes. 
Below is a construction of such pairs where $g_1$ is not even conformally equivalent to $g_2$. Start with the square flat torus $(M,g)$ obtained as the quotient of the Euclidean plane $E^2$ by the standard "square" lattice $\Gamma=\Gamma_1$ generated by  translations 
$$
T_1: z\mapsto z+1, T_i: z\mapsto z+i.
$$
(Here and below I am identifying $E^2$ with the complex plane ${\mathbb C}$ in the usual manner.)  
As my second (also flat) manifold $(N,h)$, I will take the quotient of the Euclidean plane $E^2$ by the  "nonsquare, rectangular" lattice $\Gamma_a$ generated by  translations 
$$
T_1: z\mapsto z+1, T_{ai}: z\mapsto z+ai, a>1. 
$$
It is a pleasant exercise to see that the Riemannian surfaces $(M,g), (N,h)$ are not conformally equivalent for every choice of $a>1$. (Hint: All biholomorphic automorphisms of ${\mathbb C}$ are complex-affine maps.) 
More generally, as the second lattice one can take $\Gamma_{(a,b)}$ generated by 
$$
z\mapsto z+1, z\mapsto z+ai +b,
$$
provided that $(a,b)$ is not a primitive element of ${\mathbb Z}^2$. 
On the other hand, the surfaces $(M,g), (N,h)$ are "affine-isomorphic". Namely, take the real affine transformation
$$A: x+iy \mapsto x+aiy. 
$$
Then $A\Gamma A^{-1}=\Gamma_a$ and $A$ descends to an affine diffeomorphism $f: (M,g)\to (N,h)$. Taking the pull-back Riemannian metric via $f$ on $M=M_1$ we obtain a pair of Riemannian metrics $g_1=g, g_2=f^*(h)$ on $M$ which share the same Levi-Civita connection but are not conformally-isomorphic. (It is easy to see that the connections are the same since in the local coordinates given by the universal covering $E^2\to M$ defined with the quotient by $\Gamma$, both metrics have constant coefficients, hence, zero Christoffel symbols.) 
More general examples can be obtained in a similar manner as follows. Suppose $(M,g)$ splits as a product:
$$
M=M_1\times M_2, g=h_1 + h_2,
$$
where $h_1, h_2$ are Riemannian metrics on $M_1, M_2$. Take $a>1$ and consider the new Riemannian metric 
$$
g_a= h_1 + a h_2. 
$$
It is easy to see that the metrics $g, g_a$ form an LC pair. In "most" cases, the Riemannian manifolds $(M,g), (M,g_a)$ will not be conformally-isomorphic to each other.  
On the other hand, one can prove that in a certain sense, product-manifolds are the only ones which allow for nontrivial examples:
Theorem. Suppose that $(M,g_1)$ is a Riemannian manifold such that $M$ admits another Riemannian metric $g_2$ 
such that $g_1, g_2$ form a nontrivial LC pair. Then $(M,g_1)$ is locally isometric to a product of Riemannian manifolds. 
Proof. I will need several ingredients.  The first one is the notion of the Holonomy group of an affine connection $\nabla$, $Hol_{p,\nabla}$; if $g$ is a 
Riemannian metric with Levi-Civita connection $\nabla$ then $Hol_{p,g}$ is the holonomy group of $g$. Fix a point $p\in M$ and consider the group $Hol_{p,\nabla}$ consisting of parallel transports (defined by $\nabla$ along piecewise-smooth loops based at $\nabla$). Our discussion is local, hence, I will restrict to loops contained in a totally convex neighborhood of $p$. If $\nabla$ is a Riemannian connection (for a metric $g$) then $G=Hol_{p,\nabla}$ will preserve the quadratic form $g_p$ on $T_pM$. Thus, $G$ is a relatively compact subgroup of $SO(n)=Aut(T_pM, g_p)$. This group is essentially independent of the basepoint: Holonomy groups at different base-points $p_1, p_2$ in $M$ are "conjugate" via the parallel transport along a path from $p_1$ to $p_2$.  
Next, comes a fact from elementary representation theory. Suppose that $V$ is a finite-dimensional vector space and $G< GL(V)$ is a (relatively) compact subgroup whose action on $V$ is irreducible, i.e. $V$ does not admit a nontrivial $G$-invariant direct sum decomposition $V=V_1\oplus V_2$. Then any two $G$-invariant quadratic forms $q_1, q_2$ on $V$, are scalar multiples of each other, see for instance Kyle Miller's answer here. 
Applying this result to the holonomy groups of Riemannian metrics $g_1, g_2$ on a connected manifold $M$ we obtain:
If $g_1, g_2$ have the same L-C connection $\nabla$ then either the holonomy groups $Hol_{p,\nabla}$ are  reducible for some (equivalently, every) $p\in M$ (meaning that the actions on tangent spaces are reducibe) or the metrics $g_1, g_2$ are  conformal to each other: 
$$
g_2= e^{2f}g_1, f\in C^\infty(M). 
$$ 
Another ingredient that we need is deRham's theorem: 
If $g$ is a Riemannian metric on $M$ whose holonomy is reducible, then $M$ locally splits as a Riemannian direct product, see e.g. Theorem 3.1 on p. 228 in Petersen's book "Riemannian Geometry." 
Applying this to a pair of Riemannian metrics $g_1, g_2$ with the same L-C connection, we conclude:
Either (1) $(M,g_1)$ locally splits as a Riemannian direct product or (2) the metrics $g_1, g_2$ are conformal to each other. 
Consider the case (2):
$$
g_2= e^{2f}g_1, f\in C^\infty(M). 
$$
I will prove that the function $f$ is constant, i.e. $g_1, g_2$ form a trivial LC pair. Let $grad(f)$ denote the gradient field of $f$ with respect to $g_1$. Then, since $g_1, g_2$ form an LC pair, one obtains (see Ivo Terek's answer here):
$$
X(f)Y + Y(f)X - g_1(X,Y) grad(f)=0, 
$$
for any two vector fields $X, Y$ on $M$. I claim that $grad(f)$ is identically zero on $M$, i.e. $f$ is constant. 
Indeed, consider vector fields $X=Y= grad(f)$. Then, the equation becomes 
$$
2 ||grad(f)||^2 grad(f) - ||grad(f)||^2 grad(f) = ||grad(f)||^2 grad(f)=0,
$$
i.e. $grad(f)$ is identically zero. Thus, the metric $g_2$ is a constant multiple of $g_1$. qed 
Applying this theorem inductively, one obtains the following corollary:
Corollary. Let $(M,g)$ be a Riemannian manifold which has local deRham decomposition 
$$
M=M_0\times M_1\times ... \times M_m, g= g_0 \oplus g_1\oplus ... \oplus g_m
$$
where $g_0=\delta_{ij}$ is the standard flat metric on ${\mathbb R}^k$ (and the rest of the factors are non-flat and do not split any further). Suppose that $g, h$ is an LC pair on $M$. Then (locally) $h$ has the form
$$
h= h_0 \oplus a_1 g_1\oplus ... \oplus a_m g_m
$$
where $h_0$ is a constant metric tensor on ${\mathbb R}^k$ and $a_1,...,a_m$ are certain constants. 
