Supposing a Riemannian manifold $\{M,g\}$, $\lambda g$ is also a Riemannian metric on $M$ with the constant scaling factor $\lambda>0$. The geodesics, Riemannian exponential map, Riemannian logarithm (the inverse of exponential map), and parallel transportation along the geodesic are all the same under $g$ and $\lambda g$.
Is this claim right? I think it is right. The following is my proof. The key is that Chritoffel symbols are the same.
Proof:
According to the following, Chritoffel symbols are the same: $$ \Gamma_{i j}^m=\frac{1}{2} \sum_k\left\{\frac{\partial}{\partial x_i} g_{j k}+\frac{\partial}{\partial x_j} g_{k i}-\frac{\partial}{\partial x_k} g_{i j}\right\} g^{k m} $$
According to the following two equations for geodesic and parallel transporation, we can see the equivalence of the operators under $g$ and $\lambda g$, including geodesics,exponential,logarithm,parallel transportation along a geodesic: $$ \frac{d^2 x_k}{d t^2}+\sum_{i, j} \Gamma_{i j}^k \frac{d x_i}{d t} \frac{d x_j}{d t}=0, \quad k=1, \ldots, n $$ $$ \frac{D V}{d t}=\sum_k\left\{\frac{d v^k}{d t}+\sum_{i, j} v^j \frac{d x_i}{d t} \Gamma_{i j}^k\right\} X_k=0 $$
They are not really the same but related. For example, the exponential map gives the same trajectories (the geometric geodesics) but they are traversed at a different speed. Namely, the exponential map by definition would produce unit speed geodesics, but when the metric is modified by the factor $\lambda$ this changes the speed by the same factor. The effect on other "Riemannian operators" can be deduced from their explicit form in terms of the conformal factor. For example, on a surface the Laplace-Beltrami operator would appear in conformal coordinates $f^2(x,y)(dx^2+dy^2)$ as $\frac{1}{f^2}\Delta_0$ where $\Delta_0$ is the flat Laplacian. Modifying $f$ by the factor $\lambda$ would lead to a corresponding change in the Laplace-Beltrami operator.