Consider the following bornologies $\mathbb{D},\mathbb{E}$ on the set $\mathcal{N}$ of all functions from $\mathbb{N}$ to $\mathbb{N}$:
$\mathbb{D}=\{A: \exists f\in\mathcal{N}\forall g\in A\exists m\in\mathbb{N}\forall n>m(f(n)>g(n))\}$. (Dominatable sets)
$\mathbb{E}=\{A: \exists f\in\mathcal{N}\forall g\in A\forall m\in\mathbb{N}\exists n>m(f(n)>g(n))\}$. (Escapable sets)
My question is when, set-theoretically speaking, these yield equivalent (= "bornomorphic") bounded structures on $\mathcal{N}$:
Is it consistent that $\mathfrak{b}=\mathfrak{d}$ but there is not a bijection $i:\mathcal{N}\rightarrow\mathcal{N}$ such that the $i$-image of each set in $\mathbb{D}$ is in $\mathbb{E}$ and the $i$-preimage of each set in $\mathbb{E}$ is in $\mathbb{D}$?
Here $\mathfrak{b}$ and $\mathfrak{d}$ are the cardinal characteristics corresponding to escaping(/bounding) and domination: the minimal cardinalities of sets of functions not dominated/escaped by any individual function, respectively. Clearly in order for $\mathbb{D}$ and $\mathbb{E}$ to be equivalent as bornologies we need $\mathfrak{b}=\mathfrak{d}$; I'm curious whether any additional information is captured by considering the bornological structure present.