Let $M$ denote a Riemannian manifold, $\gamma$ a geodesic of $M$ defined on $\mathbb{R}$. Let $t\_{0} \in \mathbb{R}$ and $(\alpha,\beta) \in \mathbb{R}^{2}$. I define the *reparametrized* geodesic $\tilde{\gamma} \, : \, t \, \mapsto \, \gamma(\alpha t + \beta )$. Let $v \in T\_{\tilde{\gamma}(t\_{0})}M$. $\mathrm{P}\_{t\_{0},t,\tilde{\gamma}}(v)$ denotes the parallel transport of $v$ along the geodesic $\tilde{\gamma}$ from the point $\tilde{\gamma}(t\_{0})$ to the point $\tilde{\gamma}(t)$.


I am wondering whether the following is true : $\boxed{\displaystyle \mathrm{P}\_{t\_{0},t,\tilde{\gamma}}(v) = \mathrm{P}\_{\alpha t\_{0}+\beta,\alpha t + \beta, \gamma}(v)}$


My idea is the following : let $V \, : \, t \, \mapsto \, \mathrm{P}\_{\alpha t\_{0}+\beta,\alpha t + \beta,\gamma}(v)$. $V$ is a vector field along the curve $\gamma$ and it satisfies : $V(t\_{0})=v$. In order to prove that, for all $t$, $V(t) = \mathrm{P}\_{t\_{0},t,\tilde{\gamma}}(v)$, I would need to prove that :


$$ \frac{D\_{\tilde{\gamma}}V}{dt}(t) = 0 $$


where $\frac{D\_{\tilde{\gamma}}}{dt}$ denotes the covariant derivative of $V$ along the curve $\tilde{\gamma}$. After some calculations, I find that :


$$ \frac{D\_{\tilde{\gamma}}V}{dt}(t)=\alpha \frac{D\_{\gamma}V}{dt}(\alpha t + \beta) \tag{$\star$}$$


Since $V$ is parallel along $\gamma$, I get $\displaystyle \frac{D\_{\tilde{\gamma}}V}{dt}=0$. Is that correct ?



Parallel transport along a reparametrized geodesic
This is correct. More is true: In your notation, for any continuous piecewise smooth curve
$c$ in $M$ and reparameterization $f$ one has:
$$
P\_{t\_0, f(t\_1), c} = P\_{t\_0, t\_1, c\circ f}\circ P\_{t\_0, f(t\_0), c}.
$$
See 24.1 of [here](http://www.mat.univie.ac.at/~michor/dgbook.pdf), e.g.
The proof is similar to yours. 


