Exercise: Show that a reparametrization $t → \alpha(f(t))$ of a nonconstant geodesic $\alpha$ is again a geodesic if and only if f has the form $f(t)=at+b$.
If $\alpha$ is a geodesic then it satisfies the geodesic equation. I think in Oneil in Chapter 7 that equation is given as $\alpha''=0$. This notation indicates that the covariant derivative of $\alpha'$ in the direction of $\alpha'$ is zero if I recall correctly. Let me use $\nabla_{\alpha '}\alpha' = 0$ to express that $\alpha''=0$. Notice that $\frac{d}{dt}\alpha(at+b) = a \alpha'(at+b)$ by the chain rule. Then it should be clear that $(\alpha \circ f)' = a\alpha' \circ f$ where $f(t)=at+b$. But, we know properties of the covariant derivative from earlier in the text I think, for example: for vector fields $V,W$ and function $f$ and constant $c$ we have $$ \nabla_{fV}(W) = f\nabla_V(W) \qquad \& \qquad \nabla_{V}(cW) = c\nabla_V(W)$$ hence, $$ \nabla_{(\alpha \circ f)'}(\alpha \circ f)' = \nabla_{a\alpha' \circ f}(a \alpha'\circ f) = a^2 \nabla_{\alpha' \circ f}(\alpha'\circ f) = 0$$ since we're given $\nabla_{\alpha'}(\alpha')=0$ and $(\alpha' \circ f)(t) = \alpha'(a+bt)$ and by assumption $a+bt \in \text{dom}(\alpha)$ as $f(t)=a+bt$ is a reparemetrization of $\alpha$. So, I think that is the easy direction for this bidirectional claim.
It remains to show that the only reparameterization $f$ of $\alpha$ which leaves $\alpha \circ f$ a geodesic is an affine reparametrization; $f(t)=at+b$. It seems plausible the solution here is by the existence and uniqueness theorem for DEqns applied appropriately. But, I speculate, and hence the question:
Question: can you solve the quoted exercise above? If so, is my half-solution on point, or have I forgotten a much better path.
For the best solution given in the spirit of Oneil's text I will award a bounty when the website allows. Thanks in advance for your help!
Let $\beta = \alpha \circ f$ and assume that both $\alpha$ and $\beta$ are geodesics. Then $\beta'(t) = f'(t)\alpha'(f(t))$, and taking covariant derivatives we have that $$\frac{D\beta'}{dt}(t) = f''(t)\alpha'(f(t)) + f'(t)\frac{D\alpha'}{dt}(f(t)).$$Now since both curves are geodesics, we get $f''(t)\alpha'(f(t)) = 0$. Since $\alpha$ is regular, then $f''(t) = 0$, and thus $f(t) =at+b$.
Conversely, if $\alpha$ is a geodesic and we define $\beta(t) = \alpha(at+b)$, we show that $\beta$ is a geodesic by computing $\beta'(t) = a\alpha'(at+b)$ and thus $$\frac{D\beta'}{dt}(t) = a^2 \frac{D\alpha'}{dt}(at+b) = a^2\cdot 0 = 0,$$as wanted.
On notation: if $\alpha$ is any curve and $V$ is a vector field defined along $\alpha$, $DV/dt$ is the covariant derivative of $V$, and if $V$ is defined on a neighborhood of $\alpha$, then $(DV/dt)(t) = \nabla_{\alpha'(t)}V$.