Field complete with respect to inequivalent absolute values Complete fields are known to be "robust". For instance, it is an easy exercise to show that for distinct primes $p,q$ (one of which can be the "infinite prime"), the field of $p$-adic numbers $\mathbb Q_p$  is not a subfield of $\mathbb Q_q$.
Still, I do not see an easy answer to the following question: can the same field be complete with respect to two inequivalent nontrivial absolute values? In symbols: let $K$ be a field, and let $|\cdot|_v$ and $|\cdot|_w$ be inequivalent nontrivial absolute values on $K$; is it possible that $K$ is complete with respect to both $|\cdot|_v$ and $|\cdot|_w$?
As I said above, I do not see an easy answer (though perhaps I overlooked something). But the question looks so natural that, probably, it has already being answered somewhere.
 A: For non-equivalent valued fields of characteristic 0 and same uncountable cardinality (such as $\mathbb Q_p$, $\mathbb Q_q$, $\mathbb C[[t]]$) the completions of their algebraic closures are algebraically closed fields of the same characteristic and uncountable cardinality and so are isomorphic as fields, thus can provide non-equivalent absolute values on the same (up to isomorphism) field.
A: For each prime $p$, the field $\mathbf C_p$ is isomorphic to the field $\mathbf C$ as abstract fields (i.e., they are isomorphic purely algebraically, ignoring absolute values on them) since all algebraically closed fields of the same characteristic and uncountable cardinality are isomorphic. Of course having an isomorphism in this case depends on Zorn's lemma, so field isomorphisms $\mathbf C_p \to \mathbf C$ are totally nonconstructive.
But if you're willing to accept this example then you could say $\mathbf C$ is complete with respect to infinitely many inequivalent absolute values, since it is complete with respect to its usual archimedean absolute value, while for each prime $p$ an abstract field isomorphism $\mathbf C \to \mathbf C_p$ lets us transport $|\cdot|_p$ to a non-Archimedean absolute value on $\mathbf C$ that makes $\mathbf C$ into a complete valued field where $p$ has absolute value less than $1$ and $q$ has absolute value $1$ for every prime $q \not= p$.
If you feel this example is too far out because of its very nonconstructive nature, then put some additional constraints in your question.
That the examples answering your question are rather extreme is not a surprise, since the following result shows that for down-to-earth examples, the situation you ask about is impossible.
Theorem: A field $K$ can't be locally compact with respect to two inequivalent nontrivial absolute values.
I replaced the completeness assumption with local compactness, which is a stronger condition since local compactness of a valued field implies its completeness (certainly not for general metric spaces, like $(0,1)$ with its usual metric), but not conversely, e.g., not for the fields $\mathbf C_p$.
Proof.  We're going to use the well-known result that every locally compact valued field having a nontrivial absolute value is isomorphic as a valued field to $\mathbf R$, $\mathbf C$, a finite extension  of $\mathbf Q_p$ for some prime $p$, or a Laurent series field $\mathbf F((t))$ for some finite field $\mathbf F$, each equipped with their usual kind of absolute value. EDIT: For a nontrivial non-Archimedean absolute value on a field, local compactness for the topology of that absolute value implies that absolute value's value group in $\mathbf R_{> 0}$ is discrete and the residue field is finite, since if either of those failed you could build an infinite subset of the closed unit ball with no limit point and that contradicts local compactness.
Step 1. We show the algebraic structure of the following four types of valued fields distinguishes them from each other as abstract fields: (i) $\mathbf R$, (ii) $\mathbf C$, (iii) finite extensions of $\mathbf Q_p$ for some prime $p$, (iv) Laurent series fields $\mathbf F((t))$ for some finite field $\mathbf F$.
Abstract fields of type (iv) are different from the rest due to having positive characteristic.
Abstract fields of type (i) are different from those of type (ii) and (iii) due to all positive integers being an $n$th power in the field for every positive integer $n$ and negative integers not being squares in the field.  (If $K$ is a finite extension of $\mathbf Q_p$ then the positive integer $p$ is not an $n$th power in $K$ for infinitely many $n \geq 1$.)
Abstract fields of type (ii) are different from those of type (iii) due to every nonzero integer being an $n$th power in $\mathbf C$. (I prefer to argue as much as possible in terms of numbers being powers or not, which is why I avoided the simpler method of appealing to $\mathbf C$ being algebraically closed.)
Step 2. We show that if an abstract field $K$ is locally compact with respect to a nontrivial absolute value then that absolute value is determined up to equivalence by the field structure of $K$.
By Step 1, the field structure of $K$ distinguishes the four types of fields (i), (ii), (iii), and (iv) from each other.
In cases (i) and (ii), there is only one absolute value on $\mathbf R$ and $\mathbf C$ up to equivalence that makes them locally compact. (This is part of the classification of locally compact nontrivially valued fields up to isomorphism as valued fields.)
It remains to look at fields $K$ that are isomorphic as an abstract field to a field of type (iii) or (iv). We'll show that the abstract field structure for valued fields $(K,|\cdot|)$ of type (iii) or (iv) determines the absolute value on $K$ up to equivalence.
For a valued field $(K,|\cdot|)$, the closed unit ball $\{x \in K : |x| \leq 1\}$ determines $|\cdot|$ up to equivalence: see Corollary 2.4 here. Therefore it suffices to show for each valued field $(K,|\cdot|)$ of type (iii) or (iv) that we can describe $\{x \in K : |x| \leq 1\}$ from the algebraic structure of $K$ alone. Since the valued fields $(K,|\cdot|)$ of type (iii) or (iv) are non-Archimedean, we'll write the closed unit ball in $K$ as $\mathcal O_K$.
Thanks to Hensel's lemma and finiteness of the residue field of $(K,|\cdot|)$, we can describe the group $\mathcal O_K^\times$ purely algebraically:
$$
\mathcal O_K^\times = \{x \in K^\times : x {\sf \ is \ an \ } n{\sf  th \ power \ for \ infinitely \ many \ } n \geq 1\}.
$$
In terms of $\mathcal O_K^\times$, the maximal ideal $\mathfrak m_K = \{x \in K : |x| < 1\}$ also has a purely algebraic description:
$$
\mathfrak m_K = \{x \in K^\times : x \not\in \mathcal O_K^\times {\sf \ and \ } 1+x \in \mathcal O_K^\times\}.
$$
Then $\mathcal O_K = \mathcal O_K^\times \cup \mathfrak m_K$, so $\mathcal O_K = \{x \in K : |x| \leq 1\}$ is determined by the algebraic structure of $K$.
QED
A: EDIT: I see now that my answer is not far from the second part of the answer of KConrad.
As the answers of KConrad and David Lampert explain, this is indeed possible if $K$ is algebraically closed. This is, however, essentially the only possibility, as the theorem of F. K. Schmidt tells us, see for example Theorem 4.4.1 in [1]:
Theorem (F. K. Schmidt): If $K$ satisfies Hensel's lemma with respect to two (Krull) valuations that induce distinct topologies on $K$, then it is separably closed.
This applies in particular to any two complete inequivalent nonarchimedean absolute values, and the fields that are complete with respect to archimedean absolute values are just $\mathbb{R}$ and $\mathbb{C}$.
[1] A. J. Engler and A. Prestel, Valued Fields, Springer 2005.
