Firstly Gauss law never predicts that there will be no $\vec E$ inside the conductor , the fact that $\vec E$ is $0$ comes from the fact that we are dealing with electrostatics . What gauss law predicts is that $Q_{inside}$ a conductor will be $0$ under a very special condition/state called electrostatic.
Now what is electrostatics ?
It is a condition achieved , a physical state such that no charge anywhere is moving.
Using this argument to say $\vec E=0$ in a conductor . We reason as follows , we know that any conductor by definition has something has de-localised electrons / charge carriers which are free to flow inside the conductor at least . Now if there's any $\vec E$ inside the conductor , we can surely say then the charges must move . But we are assuming there is electrostatic condition , means no charges are moving . Thus $\vec E$ must be $0$ , because had it not been $0$ we could have never achieved electrostatic condition .
Now the Gauss' law part , as $\vec E$ is 0 , so Flux $\phi=0$ through out gaussian surface by definition of flux and thus it means by Gauss' law that charge contained inside our Gaussian surface is $0$ ,
So we proved that in electrostatic conditions , charge cannot reside inside a conductor using
1. A physical assumption &
2. A law of nature .
Now dealing with your question , assuming electrostatic condition has been achieved , there must be no $\vec E$ inside the conductor , I am assuming not a hollow sphere . Thus net charge inside any gaussian surface you imagine is $0$ .
Now imagining this physically , suppose the whole sphere is made up of very very thin infinitesimal shells , you can assume condition to be like this in every infinitesimal shell (assume these to bespheres).
But you know net charge of an isolated system is conserved ,
so on the outermost surface there will be net charge = the charge you put inside in the first place.
Equal in terms of magnitude and sign .
And it'll also be uniformly distributed . Why ?
Because of the symmetric nature of the sphere , suppose you are the charge , you'll see that all points are adzactly equal to each other in every respect , orientation etc. so you'll distribute same way at each point and also if you calculate , in general if you reason that charge will be more here , I can reason same way for any other point on the sphere , you'll see that only this symmetric distribution here helps us to achieve Electrostatic Conditions .
Best Answer
Yes, the charges rearrange to that the total field inside the conductor is zero---that is, the sum of the fields due to $q_1$ and $q_2$ will vanish. No, the total amount of charge inside the conductor does not change, but it might happen that there's more positive charge here, more negative charge there, and so on, but always in a way that keeps the total amount of charge constant.