Let $X$ and $Y$ be Banach spaces such that $\mathcal{B}(X)$ is linearly isomorphic to $\mathcal{B}(Y)$ (where $\mathcal{B}(\cdot)$ denotes the algebra of bounded linear operators). Must it always be the case that $X$ is therefore isomorphic to $Y$?
I suspect this is not true, but I can't immediately see how to produce a counter-example. Surely, whether it is true or false, this must be well-known.
Ben, the answer is no.
Note that if $X$ is reflexive, then $B(X)$ is isometric to $B(X^*)$ via $T\mapsto T^*$. Note that this map is an anti-isomorphism of Banach algebras.
As for less trivial examples, $B(\ell_p)$ is Banach-space isomorphic to $B(L_p)$ as well to $B(X)$ for any other separable, infinite-dimensional $\mathscr{L}_p$-space and $\ell_p$ is not isomorphic to $L_p$ unless $p=2$. This was observed by Arias and Farmer but I suppose it had been known before to the big shots in the field.