action - Why does reparameterisation invariance lead to gauge-fixing?

In Becker, Becker and Schwarz, the point particle action is given in terms of an auxiliary field $e(\tau)$ as:

$$\begin{align} \tilde{S}_0 = \frac{1}{2}\int \,d\tau \left(e^{-1}\dot{X}^2 - m^2e\right) \end{align}$$

It is then shown that under infinitesimal reparametrizations of $\tau$, the action is unchanged. This allows us to pick a gauge, in particular $e(\tau) = 1$.

I'm not sure I'm understanding this right, but I have a few issues with this.

1. 
   

   Doesn't this assume that $e$ takes the value $1$ somewhere?

   
   


2. 
   

   Although $\tilde{S}_0$ (sorry, not $e(\tau)$) may be reparametrization invariant, I don't see how you could pick a reparametrization that can leave $e$ constant. Such a reparametrization $\tau'(\tau)$ would need to map all $\tau$ to a constant, but then $\frac{d\tau'}{d\tau} = 0$, which can't be the case.

   
   



3. 
   

   Invariance is only under infinitesimal transformations. This is related to 2: how do we know that an infinitesimal reparametrisation could make $e$ constant?