Here is a thought on the first question. What you need to know (at least to get an algebraic space; I'll let others be more careful than I if you want a scheme) is how large n must be to ensure that an automorphism of a smooth genus g curve X which fixes n points must be the identity. Let G be the cyclic group generated by this automorphism: then the map X -> X/G is totally ramified at your n fixed points. So by Riemann-Hurwitz, g(X) is at least -2|G| + n(|G|-1). If G is nontrivial, in other words, g is at least n-4. So I think g+5 marked points should be enough. That this is necessary can be seen by taking g=2; on M_{2,6} you'll have a bunch of loci with an extra involution, parametrizing curves whose marked points are precisely the Weierstrass points.

Here is a thought on the first question. What you need to know (at least to get an algebraic space; I'll let others be more careful than I if you want a scheme) is how large n must be to ensure that an automorphism of a smooth genus g curve X which fixes n points must be the identity. Let G be the cyclic group generated by this automorphism: then the map X -> X/G is totally ramified at your n fixed points. So by Riemann-Hurwitz, g(X) [NO, 2g(X)-2, THANKS, BJORN) is at least -2|G| + n(|G|-1). If G is nontrivial, in other words, g is at least n-4 [NO, 2g+2, THANKS, BJORN]. So I think g+5 [NO, 2g+3, THANKS, BJORN] marked points should be enough. That this is necessary can be seen by taking g=2; on M_{2,6} you'll have a bunch of loci with an extra involution, parametrizing curves whose marked points are precisely the Weierstrass points.

[NO MORE LATE-NIGHT RIEMANN-HURWITZ: THANKS TO BJORN FOR CORRECTING THE ERRORS]