Hereinafter we will assume S~ and S are connected spaces. If p is the projection from S~ onto S, an automorphism can be viewed as a lift of the projection up to S~. Two such lifts are identical, or everywhere distinct. Select a base point upstairs and down, such as x~/x, and the image of x~, another point in the fiber of x, completely determines the automorphism. Thus the automorphism group of S~ becomes a subgroup of the symmetric group on the points in the fiber of x.
Let y~ be another point in the fiber of x. Is there an automorphism taking x~ to y~? The two spaces, S~ and S~, are connected, and inject the same fundamental group into S, so an automorphism exists. There exists an automorphism for each y~, and the size of the automorphism group equals the cardinality of the fiber of x.
For example, the automorphism group of R1 over S1 is Z. The real line can be shifted left or right by any integer.
When Sn is a cover for projective space Pn, the automorphism group is Z2, identity and reflection.
Let a path m~ in S~ connect a1 to a2, another point in the fiber of x. Of course m~ projects to a loop m downstairs. Assume m is in the normalizer of G. In other words, mG/m = G.
Let l1 be a loop based at a1. Project l1 down to a loop l based at x, then lift l to a path l2 starting at a2. We will show l2 is a loop.
Lift ml/m up to a path that starts at a1, goes to a2, travels along l2, and backs up along a path equal to, or parallel to, m~. Since ml/m is another element of G, it is the image of a loop based at a1. Thus the lift returns to a1, and l2 must start and end at a2.
Now for the converse. Assume every loop l1 projects and lifts to a loop l2. For every l in G, ml/m lifts to a loop based at a1. Thus ml/m lies in G, and m is in the normalizer of G.
This is directly applicable when h is an automorphism taking a1 to a2. Let m~ be a path from a1 to a2, projecting to a loop m downstairs. Since h is continuous, it maps loops to loops. Thus h(l1) = l2. Yet h also respects projection, whence l1 and l2 lead to the same loop l downstairs. The lift of the projection of a loop is always a loop, and that means m is in the normalizer of G.
Let n~ be another path from a2 back to a1. Apply m~ backwards, from a1 to a2, then n~, and find a loop upstairs, which becomes an element of G downstairs. Thus m inverse followed by n belongs to G. Multiply by m on the left, and m and n represent the same left coset of G in its normalizer. We have constructed a map, which I will call f, from the automorphisms of S~/S into the left cosets of G within its normalizer.
By construction, G is normal within its normalizer, hence there is no significant distinction between left and right cosets. We simply have cosets of G, which are also elements of the quotient group n(G)/G.
Consider the composition of two automorphisms h and j. Let h map a1 to a2 as above, with a path m~ connecting a2 to a1. Let j carry a1 to a3, with a path n~. Being continuous, j carries the entire path m~ to another path from a4 to a3. Thus a4 = j(h(a1)).
when applying f, any path will do, so travel along j(m~) to a3, then along n~. Since j respects projection, both j(m~) and m~ map to m downstairs. This is followed by n. Therefore the composition of automorphisms leads to the concatenation of loops representing cosets of G. In other words, f is a group homomorphism.
Injective is not hard to prove. Let m~ from a2 to a1 map to a loop in G. Every loop in G comes from a loop in S~, hence m~ actually defines a loop, and the automorphism is trivial. Thus f is a monomorphism.
Finally, show f is surjective, giving an isomorphism. Let m be the inverse of a loop that represents a coset of G in its normalizer. Lift m up to a path that runs from a2 back to a1. We already said there is a lift, an automorphism, carrying a1 to a2, and that automorphism leads to m.
In summary, the automorphism group of S~/S is equal to the quotient group n(G)/G, where G is the image of π(S~) in S.
As an example, review the covering space we built for the figure 8. A free group on infinitely many generators embeds in the free group on A and B, and this embedding has to be normal. The fiber is the set of integers, the powers of A, and the automorphisms shift the real line, with its dangling loops, left or right.
You can verify this algebraically. The embedded group G in the figure 8 is generated by AnB/An for all values of n. A word belongs to G iff it presents an equal number of A's and A inverses. Conjugate such a word by A or B and the criterion still holds, hence G is a normal subgroup. Finally, the cosets of G, the elements of the quotient group, are the positive and negative powers of A.
Let S2 act as a covering space for P2 in the obvious way. Each point in projective space now has two points in its fiber. The automorphism group of S2/P2 is Z2. We already showed the sphere is simply connected, so it's fundamental group is trivial. Embed this in P2, and G is trivial. Therefore π(P2) = Z2. This holds for all dimensions beyond 1.