Let $C$ be a convex subset of $\mathbb{R}^{α}$ for an ordinal $α$. Call $C$ nice iff for every linear map $λ : \mathbb{R} \to \mathbb{R}^α$, the preimage $λ^{-1}(C) ⊆ \mathbb{R}$ is bounded.
An example of a nice set is the Cartesian product of intervals $[-1, 1]^α$. I believe that if $α$ is finite, then for any nice subset of $\mathbb{R}^α$, there is an isomorphism $\phi : \mathbb{R}^α \to \mathbb{R}^α$ such that $\phi(C) ⊆ [-1, 1]^α$. Is this also true for infinite $α$?
The answer is no. Let $α = ω + 1$. I'll write elements of $\mathbb{R}^{ω+1}$ as pairs of an element of $\mathbb{R}^\omega$ and an element of $\mathbb{R}$. As this answer (and this one) discusses, there is an unbounded linear function $f : [0, 1]^ω \to \mathbb{R}$. Let $$ C = \big\{ (x, y) ∈ \mathbb{R}^{ω + 1} \; \big| \; x ∈ [0, 1]^ω, \; 0 ≤ y ≤ f(x) \big\} $$
$C$ is convex. If $(x, y)$ and $(x', y')$ are in $C$, then we have \begin{align} 0 & ≤ λy + (1 - λ)y' \\ & ≤ λf(x) + (1 - λ)f(x') \\ & = f(λx + (1 - λ)x') \end{align}
$C$ is nice. Let $λ : \mathbb{R} \to \mathbb{R}^{ω + 1}$ be a linear map. Let $(x, y) = λ(1) ∈ \mathbb{R}^{ω + 1}$. If $x = 0$, then $f(x) = 0$, so $y = 0$, so $λ^{-1}(C) = \{0\}$. Otherwise, for $a > (\sup_i x_i)^{-1}$, we have $(λ(a))_i = a x_i > 1$, and thus $λ(a) ∉ C$. So $λ^{-1}(C)$ is bounded.
But no isomorphism $\phi$ maps $C$ into the cube. For any isomorphism $\phi$, and any $x ∈ [0, 1]^ω$, we have $$ \phi(x, f(x)) = \phi(x, 0) + f(x) \, \phi(0, 1) $$ If the first term is always in the cube, then since $f(x)$ is unbounded (and $\phi(0, 1) ≠ 0$) this must land outside of the cube for some $x$.