Q1: Definately not always. More like "almost never". If the automorphism extends to , then the bundle would embed in . is the bundle over with fiber and monodromy . The most-commonly used obstructions to embedding in this case are things like the Alexander polynomial, and Milnor signatures.
I don't see where the metric on plays a role for this.
If you want to see automorphisms that extend (and do not extend) for your Q1, take a look at my arXiv paper. You'll also find some references to several Jonathan Hillman papers that explore such obstructions.
In the case that your surface is unknotted -- bounding handlebodies on both sides (thinking of ) then the automorphisms of that extend in this case are well-known. They're called the mapping class group of a Heegaard splitting of . It's an infinite group. Generators are known for it (if I recall, they're the automorphisms induced by handle slides) but off the top of my head I'm not sure how much is known about the structure of the group. Do a little Googling on "mapping class group of a Heegaard splitting of S^3" and you should start finding relevant material.
To respond to your 2nd edit, if the co-dimension is high enough all automorphisms extend. This is a theorem of Hassler Whitney's. The basic idea is this, let be an automorphism. Let be any embedding. So you have two embeddings and of in . Any two maps are isotopic provided the co-dimension is large enough suffices, for example. So isotope your standard inclusion to the one pre-composed with . The Isotopy Extension Theorem gives you the result.
For example, if is a Heegaard splitting / the surface is unknotted, (or ) and you have an automorphism a neccessary and sufficient condition for to extend to (or a side-preserving automorphism of the pair in the case) is that if is a curve that bounds a disc on either the inside or outside of respectively, then bounds a disc on the inside or outside of respectively (here I'm using inside/outside re the Jordan-Brouwer separation theorem). Since the fundamental group of the complement is just a free product of infinite cyclic groups this is something that can be checked rather easily provided you know the map well enough.
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