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I would like to know what is $Hom_{Groups}(K^*,K^*)$, at least in the case $K$ is a complete non-archimedean (valued) field. Is this $\mathbb{Z}$?

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up vote 3 down vote accepted

Much bigger than $\mathbb Z$, at least if (as is standard) you mean by “$K^*$” the set of nonzero elements of $K$. (Your title suggests something different.)

I will say something about the automorphisms of a local field of mixed characteristic, i.e. a finite extension of $\mathbb Q_p$. The question is certainly much more difficult if you accept discontinuous automorphisms, and I will not consider such. What you need to do is decompose $K^*$ as direct sum of simpler groups, and you find that $$ K^*\cong \mathbb Z\oplus W\oplus(\mathbb Z_p)^n\>, $$ where $W$ is the (finite) group of roots of unity of $K$, and $n=[K\colon\mathbb Q_p]$. This decomposition is not unique. Now to see all automorphisms of this group you need to look not only at the automorphisms of the summands but also at the homomorphisms from one summand to another.

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