Sequences are Cauchy depending on the norm?

Assume that we have two Banach spaces $B_1, B_2$ with their underlying sets being identical. Is it possible that a Cauchy sequence in one of the spaces would fail to be Cauchy in the other? I am assuming that it is possible since the Cauchy test directly depends on the norm. However, this seems kind of unnatural, will someone please clarify.

Here are some details to add to Jonas' answer in the link of my comment above.

Let $V$ be a vector space of Hamel dimension $c$. Let $T$ be a linear bijection from $V$ to $\ell_1$ and let $S$ be a linear bijection from $V$ to $\ell_2$. This can be done since any two separable infinite dimensional Banach spaces have Hamel dimension $c$.

Define two norms on $V$ via $\Vert x\Vert_T=\Vert Tx\Vert_{\ell_1}$ and $\Vert x\Vert_S=\Vert S x\Vert_{\ell_2}$. Then $(V,\Vert\cdot\Vert_S)$ is isomorphic to $\ell_2$ and $(V,\Vert\cdot\Vert_T)$ is isomorphic to $\ell_1$. In particular, these two norms are complete.

Since $\ell_1$ is not isomorphic to $\ell_2$, it follows that these two norms are not equivalent. Per Julien's answer, these two norms have different Cauchy sequences.

Edit: here is a slightly different approach for a counterexample, without reference to any exterior Banach space. We claim that for every infinite-dimensional Banach space $(X,\|\cdot\|)$, there exists another complete norm $\|\cdot \|'$ on $X$ which is not equivalent to the original norm and therefore has different Cauchy sequences.

Let $\{x_j\,;\, j\in J\}$ be a Hamel basis for $X$. Without loss of generality, we can assume that $\|x_j\|=1$ for every $j\in J$. We know by Baire that $J$ is not countable. But all we need is that there exists $\mathbb{N}^*\simeq J_0\subseteq J$. Now consider the linear operator $T:X\longrightarrow X$ defined on the basis by $Tx_n:=nx_n$ for all $n\in\mathbb{N}^*\simeq J_0$ and $T_jx_j:=x_j$ for $j\in J\setminus J_0$. By construction, $T$ is invertible and unbounded.

Define the new norm by $\|x\|':=\|Tx\|$. Then $T:(X,\|\cdot\|')\longrightarrow (X,\|\cdot\|)$ is an isometric isomorphism whence $(X,\|\cdot\|')$ is complete. But $\mbox{Id}:(X,\|\cdot\|)\longrightarrow (X,\|\cdot\|')$ is not bounded since $T$ is not bounded by construction. So the two norms are not equivalent.

Note that it is easy to adapt the argument to construct infinitely many complete pairwise non-equivalent norms on $X$. Just because $J$ contains $\mathbb{N}\simeq \mathbb{N}\times \mathbb{N}$.

In a Banach space, a sequence is Cauchy if and only if it converges. So you are asking when two complete norms $\|x\|_1$ and $\|x\|_2$ on a given space $X$ yield the same converging sequences.

Since these are metric spaces, the topologies are sequential. So you are asking when two distinct (complete) norms define the same topology. That is, when $$ \mbox{Id}:(X,\|\cdot\|_1)\longrightarrow (X,\|\cdot\|_2) $$ is a homeomorphism. Since this is an invertible linear map, this is equivalent to boundedness in both directions. That is, to the existence of positive constants $A,B$ such that $$ A\|x\|_1\leq \|x\|_2\leq B\|x\|_1\quad \forall x\in X. $$ We say that the norms are equivalent.

But by Banach isomorphism theorem (or the open mapping if you prefer), it is sufficient that one side of the inequality hold. In other terms: if every Cauchy sequence for one norm is Cauchy for the other, the norms are equivalent and the converse holds.

Example: on a finite-dimensional real or complex vector space, all the norms are equivalent and complete.

Example: $\|f\|_\infty=\sup_{[0,1]}|f(x)|$ and $\|f\|=\|f\|_\infty+|f(0)|$ are two equivalent complete norms on $C^0([0,1])$.

Oops, I forgot to answer the question...

Counterexample: David Mitra's counterexample is perfect. See this article for the existence of spaces with infinitely many complete norms such that the resulting Banach spaces are pairwise non-isomorphic (in particular, the norms are pairwise non-equivalent, but this result is much stronger).