In models of Zermelo-Fraenkel without the axiom of choice (AC), it is consistent to say that there exist vector spaces $V$ with bases of different cardinalities, say $B$ and $B'$. Let $V$ be a vector space over the field $k$.
Considere the general linear group $G = \mathrm{GL}(V)$ of $V$ (which consists of invertible linear transformations). Suppose $\vert B \vert = \omega$, and $\vert B' \vert = \omega'$ (where both $\omega$ and $\omega'$ are not finite). If we represent $G$ relative to the base $B$, we obtain $\mathrm{GL}(\omega,k)$ and if we do it relative to $B'$, we obtain $\mathrm{GL}(\omega',k)$.
I might be abusing notation here, but if not: I guess this means that in this model, $\mathrm{GL}(\omega,k) \cong \mathrm{GL}(\omega',k)$, while $\omega \ne \omega'$?