Let $X\to \mathbf{P}^1$ be a branched cover of the complex projective line, where $X$ is a compact connected Riemann surface. Let $G=\mathrm{Aut}(Y/\mathbf{P}^1)$.
Question 1. Could somebody provide some examples of branched covers $X\to \mathbf{P}^1$ for which $G= (0)$?
One could construct a compact connected Riemann surface without automorphisms to this end.
On
Here is another perspective. The category of compact Riemann surfaces is (anti-)equivalent to the category of finitely generated fields over $\mathbb C$ of transcendence degree $1$, given by $C\mapsto k(C)$, the field of rational functions [Reference: Appendix of Mumford's Red Book].
Now, the data of a cover $C\to\mathbb P^1$ is equivalent to a field extension of a field extension $L/\mathbb C(x)$, and the automorphisms of $C\to\mathbb P^1$ are exactly the automorphisms of $L/\mathbb C(x)$.
For example, let $L/\mathbb C(x)$ be a degree $3$ non-Galois extension, such as $L=\mathbb C(x)[y]/(y^3-xy-1)$. Then, $L/\mathbb C(x)$ has no automorphisms. The corresponding curve $C$ is the elliptic curve $E\colon y^3-xyz-z^3=0$ in projective coordinates $[x:y:z]$.
As you suggest, it is enough to prove the existence of a compact Riemann surface $X$ with only the identity as automorphism .
Indeed, $X$ has a non constant meromorphic function $f$ and by considering it as a ramified covering $f:X\to \mathbb P^1$ we get our example .
Actually examples are plenty: a generic compact Riemann surface of genus $g\geq 3$ has no automorphism besides the identity . This is a theorem proved by by Baily in 1961.
You can get a feel for that theorem by going to this MathOverflow question, to which Torsten Ekedahl (whom we sadly lost little ago) gave a great answer.