Is every infinite-dimensional Banach space coarsely equivalent to its subspaces of codimension 1?

Question

Gowers in his solution to the Banach hyperplane problem constructed a Banach space which is not linearly isomorphic to any of its codimension one subspaces.

However, how about much weaker notion of `coarse equivalence'? That is, let $X$ be an infinite-dimensional Banach space and let $Y$ be a codimension one closed subspace, i.e. a Banach space such that $Y\oplus\mathbb{R}\simeq X$. Are $Y$ and $X$ coarsely equivalent?

Recall that two Banach spaces $X$ and $Y$ (or in general metric or just coarse spaces) are coarsely equivalent if there exist

(1) two non-decreasing functions $\phi,\psi: [0,\infty)\rightarrow [0,\infty)$ with $\lim_{r\to\infty} \phi(r)\to\infty$;

(2) function $f:X\rightarrow Y$;

(3) constant $K>0$;

such that we have for all $x,y\in X$ $$\phi\big(\|x-y\|_X\big)\leq \|f(x)-f(y)\|_Y\leq \psi\big(\|x-y\|_X\big),$$ and the image $f[X]$ is $K$-cobounded, i.e. for every $y\in Y$ there is $x\in X$ with $\|y-f(x)\|_Y\leq K$.