Let E/K and F/K be two field extensions of K. Note that K contains the base field generated by 1, either Q or Zp. Recall that E and F are both vector spaces over K. Let S be the set of K module homomorphisms from E into F. These are also known as linear transformations. Each function in S can be scaled by an element of K, and functions in S can be added together to get another function in S. In other words, S is a K vector space.
Let T be the functions in S that are nontrivial field homomorphisms. In other words, the functions in T preserve multiplication. Since each function c in T is nontrivial, we must have c(1) = 1. Since c is a K module homomorphism, as well as a field homomorphism, c(K) = K. Every function in T fixes K. There may be other field homomorphisms that move parts of K, but these are not included in S, and not included in T.
You can add or scale the functions in T, but you probably won't get another function in T. If K is the real numbers and c is complex conjugation, 2c is not a field homomorphism, because 2c(1) = 2, whereas a field homomorphism always maps 1 to 1.
Suppose there is a linear combination of field homomorphisms in T that produces 0. We usually take coefficients from K, but let's generalize this to include coefficients from F. Since there is some linear combination that produces 0, with coefficients in K, there are linear combinations that lead to 0 with coefficients in F. We may as well select the shortest example.
Let the sum of aici give the 0 homomorphism, where each ci is a field homomorphism in T, and each ai is a coefficient taken from F. This is combination 1. There are at least two homomorphisms in combination 1, else c1 or a1 is trivially 0.
If c1 and c2 agree everywhere, they are in fact the same function, so select y such that c1(y) ≠ c2(y). If c1(y) is 0, swap the roles of c1 and c2. Now we know c1(y) is nonzero. In the following, y is held constant.
For any x in E, do the following. Apply our shortest linear combination to xy. Thus the sum of aici(xy) = 0. Pull each ci(y) out and make it part of the coefficient. This gives another linear combination of homomorphisms, combination 2, that still maps every x to 0.
Next, multiply combination 1 through by c1(y). This is combination 3, a scaled version of combination 1. For every x, the sum of aici(x)c1(y) = 0. Subtract combination 2 from conbination 3, giving combination 4. The first term, a1c1(xy) drops out. The second term is a2(c1(y)-c2(y))c2(x). Since c1 and c2 disagree on y, the coefficient is nonzero, and the second term remains. Combination 4 is a shorter linear combination of homomorphisms that yields 0. This is a contradiction, therefore all field homomorphisms are linearly independent.
Since T, the set of field homomorphisms, is an independent set, its size is bounded by the dimension of S, as a vector space over K. Pick a basis for E, any basis. Since every homomorphism in S determines, and is determined by, the image of this basis in F, the dimension of S is the same as the dimension of E as a K vector space. If E/K is an extension of dimension 5, there are at most 5 field homomorphisms from E/K into F/K, fixing K. Of course F/K could be the same as E/K, whence the field monomorphisms become automorphisms. There are at most 5 automorphisms of E fixing K.