In the context of the Polyakov action $$ S_P = \frac{1}{2} \int \mathrm{d}\tau \, e(\tau) \left(\frac{1}{e^2(\tau)}\left(\frac{\mathrm{d} X^\mu(\tau)}{\mathrm{d} \tau}\right)^2 - m^2\right), $$ where $\tau$ is an arbitrary parameter, and $X^\mu(\tau)$ and $e(\tau)$ are independent degrees of freedom, I recently read that invariance under local diffeomorphisms $\tau \to \tilde{\tau} - \epsilon(\tau)$ requires $X^\mu(\tau)$ and $e(\tau)$ to transform as $$ X^\mu(\tau) \to X^\mu(\tau) + \frac{\mathrm{d} X^\mu(\tau)}{\mathrm{d} \tau} \epsilon(\tau),\\ e(\tau) \to e(\tau) + \frac{\mathrm{d}}{\mathrm{d} \tau}\bigl(e(\tau) \epsilon(\tau)\bigr). $$ My question is, why do we require invariance under local diffeomorphisms in the first place. What does it represent physically?

In the context of the Polyakov action, the action for a relativistic point particle $$ S_P = \frac{1}{2} \int \mathrm{d}\tau \, e(\tau) \left(\frac{1}{e^2(\tau)}\left(\frac{\mathrm{d} X^\mu(\tau)}{\mathrm{d} \tau}\right)^2 - m^2\right), $$ where $\tau$ is an arbitrary parameter, and $X^\mu(\tau)$ and $e(\tau)$ are independent degrees of freedom, I recently read that invariance under local diffeomorphisms $\tau \to \tilde{\tau} - \epsilon(\tau)$ requires $X^\mu(\tau)$ and $e(\tau)$ to transform as $$ X^\mu(\tau) \to X^\mu(\tau) + \frac{\mathrm{d} X^\mu(\tau)}{\mathrm{d} \tau} \epsilon(\tau),\\ e(\tau) \to e(\tau) + \frac{\mathrm{d}}{\mathrm{d} \tau}\bigl(e(\tau) \epsilon(\tau)\bigr). $$ My question is, why do we require invariance under local diffeomorphisms in the first place. What does it represent physically?