I am thinking about the following question:


Let $X$ be a Banach space, say separable, e.g., $l\_p$ or $c\_0$. 
When can I say that there exist inequivalent complete norms on $X$?



inequivalent norms
First, let me mention the result of Mackey: *if $X$ is an infinite-dimensional Banach space, then its Hamel dimension* (= dimension as abstract vector space, over $\mathbf{R}$ or $\mathbf{C}$) *is $\ge c=2^{\aleph\_0}$, with equality if $X$ is separable*.


In the separable case, $X$ has cardinal $\le c$ so the upper bound is clear. The lower bound is trivial under CH (continuum hypothesis) using Baire. In general (i.e., not assuming CH), here's a proof of $\ge$ due to Lacey (mentioned at MathSE [here](https://math.stackexchange.com/a/141543/) and [here](https://math.stackexchange.com/q/1899365/35400)). Using Hahn-Banach, there exist sequences $(x\_n)$ and $(\ell\_n)$ in $X$ and $X^\*$ (the topological dual) such that $\ell\_m(x\_n)=\delta\_{m,n}$ for all $m,n$ (Kronecker's $\delta$). It follows that for every $n$, $x\_n\notin\overline{\mathrm{span}\{x\_m:m\neq n\}}$. Now use the existence of a family $(J\_t)\_t$ of cardinal $c$ of infinite subsets of $\mathbf{N}$ with pairwise finite intersection. [To obtain this, take the countable set as the set of vertices of an infinite regular rooted binary tree, and the subsets as the set of infinite geodesic rays starting from the root.] Then the family $(y\_t)$ defined by $y\_t=\sum\_{n\in J\_t}2^{-n}x\_n$ is linearly independent by an easy argument.


(Of course in $\ell^p$ or $c\_0$ one can produce these elements without using Hahn-Banach, namely $x\_n$ can be chosen as Dirac at $n$.)




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Next, the assertion is that *for a vector space $V$ of infinite dimension $\alpha$ over a field $K$ of cardinal $\beta$ with $\beta\le 2^\alpha$, the cardinal of $\mathrm{GL}(V)$ is $\beta^\alpha$, and in particular equals $2^\alpha$ if $\beta\le 2^\alpha$ (which is automatic if the field has cardinal $\beta\le c$).*


For the latter fact, just use that $\beta^\alpha\le (2^\alpha)^\alpha=2^{\alpha\times\alpha}=2^\alpha$.


For the upper bound, do it for $\mathrm{End}(V)$, which is in bijection with the set of maps $\alpha\to V$, i.e., has cardinal $\beta^\alpha$. 


For the lower bound, splitting a basis, write $V=W\oplus W$ with $W$ isomorphic to $V$. Then every $u\in\mathrm{End}(W)$ induces $f\_u:(x,y)\mapsto (x+u(y),y)$, with $f\in\mathrm{GL}(V)$, these provide $\beta^\alpha$ automorphisms. (Alternatively we already get $2^\alpha$ automorphisms by permuting basis elements— say gather elements by pairs and for any subset of this set of $\alpha$ pairs transpose them and leave identity elsewhere.)




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Combining, we obtain that for $X$ infinite-dimensional separable Banach space, the automorphism group $G\_X$ of $X$ as abstract vector space (over $\mathbf{R}$ or $\mathbf{C}$) has cardinal $2^c$.


If $H\_X\subset G\_X$ is the subgroup of (bi)continuous automorphisms then $H\_X$ has cardinal $\le c$ and hence $G\_X/H\_X$ has cardinal $2^c$, so the $g\cdot (\|\cdot\|)$, when $g$ ranges over $G\_X/H\_X$, yield $2^c$ pairwise non-equivalent norms on $X$.


One can elaborate using cardinal invariants about the non-separable case, but for free we get $2^c$ non-equivalent norms in $X$ without assumption: choose an infinite-dimensional separable subspace $Y$, and define $G$ as the group of abstract linear automorphisms of $X$ that preserve $Y$, and $H$ its subgroup of bicontinuous automorphisms: then looking at the action on $Y$ shows that $H$ has index $\ge 2^c$ in $G$, and again the $g\cdot (\|\cdot\|)$ for $g$ ranging over $G/H$ are pairwise non-equivalent norms on $X$ (they're non-equivalent since they're non-equivalent in restriction to $Y$).


