Suppose that $M$ is a smooth manifold enodwed with a connection, $\gamma :I \to M$ is a smooth curve, and $Y \in \mathfrak{X}(\gamma)$. Prove that if $Y$ is parallel along $\gamma$, then it is parallel along every reparametrization of $\gamma$.
So if we suppose that $Y$ is parallel along $\gamma$, then $D_tY= 0$ which in coordinates says that $$D_tY(t)=(\dot{Y}^k(t)+ \dot{\gamma}^i(t)Y^j\Gamma_{ij}^k(\gamma(t)))\partial_k\vert_{\gamma(t)} = 0.$$ Now if $\widetilde{\gamma}$ is a reparametrzation of $\gamma$, then $\widetilde{\gamma} = \gamma\circ \varphi: I' \to M$ where $\varphi:I'\to I$ is a diffeomorphism. So what we would want to get is $$D_tY(t)=(\dot{Y}^k(t)+ \dot{\widetilde\gamma}^i(t)Y^j\Gamma_{ij}^k(\widetilde\gamma(t)))\partial_k\vert_{\widetilde\gamma(t)} = 0$$ and I'm thinking this is immediate from this equaling $$D_tY(t)=\left(\dot{Y}^k(t)+ \dot{\gamma(\varphi(t))}^iY^j\Gamma_{ij}^k(\gamma(\varphi(t)))\right)\partial_k\vert_{\gamma(\varphi(t))}$$ and having the above parallel condition on $\gamma$. Am I making some error here as it does not seem that I actually had to do anything?
You were almost there but not quite. Remember that $D_t Y$ is just a notation for $D_{\dot{\gamma}(t)} Y$. So what you actually want to get is $D_{\tilde{t}} Y=(\dot{Y^k}(\tilde{t})+\Gamma_{i\,j}^kY^j(\tilde{t})\tilde{\gamma^j} )g_k$. This is not quite equal do $D_t Y$; but it is proportional to it. By what factor? By the one given by the chainrule, $\frac{d\tilde{t}}{dt}$, which appears on both terms, one from the derivative of the field, one from the derivative of the curve, and so it is just factored out. So, under reparametrisation, we have $D_{\tilde{t}} Y=\frac{d\tilde{t}}{dt} D_t Y$, which is even suggestive of some Leibniz type mnemonic for the chainrule. In particular if one is zero, the other one is null as well.