A local field (to my knowledge) is a field equipped with a non-trivial absolute value that makes it a locally compact topological space. From this, one derives that the local fields are $\Bbb R,\Bbb C,$ finite extensions of $\Bbb Q_p,$ and $\Bbb F_q((x))$ with the usual absolute values/valuations.
If $K$ is a local field with respect to some absolute value $|\cdot|$, can there be a further, non-equivalent absolute value on $K$ making it again local? I have the strong suspicion that the answer is "no", but would like to know precisely how this can be proved.
A field that is complete with respect to two non-equivalent absolute values is separably closed. So if $K\ne \Bbb C$, the answer to your question is no. For $K=\Bbb C$, one can also take other absolute values: Take an automorphism $\sigma$ of $\Bbb C$ which is not the identity or complex conjugation. Then $\lVert x\rVert:=|\sigma(x)|$ defines another absolute value of $\Bbb C$ for which $\Bbb C$ is again a local field (as it will be isomorphic to the usual $\Bbb C$ via $\sigma$).
The idea of the proof of the first statement is roughly the following: Take an irreducible separable polynomial $g$ over $K$. There is a polynomial $h$ of the same degree that splits completely over $K$. Then using the approximation theorem choose a polynomial $f$ (of the same degree) which is very close to $g$ w.r.t. to the first absolute value and very close to $h$ w.r.t. the other. Applying Krasner's lemma for both absolute values then shows that $f$ must be both irreducible and split completely over $K$, which forces $\deg g=\deg f=1$, so $K$ is separably closed.