It seems common amongst logicians to think of "truth" as being relative to a particular structure. Consider, for instance, the first-order theory of groups. The sentence $\forall x\forall y(x\cdot y=y\cdot x)$ is neither provable nor disprovable from the group axioms. However, rather than saying this sentence is neither true nor false, we would simply say that it is true for some groups (namely, the abelian ones), and false for others. The situation seems similar for most set-theoretical issues – for instance, while saying the continuum hypothesis is true without any qualification is clearly controversial, all set theorists can agree that the continuum hypothesis is true in some models of set theory, and not in others.
However, the situation feels a little different when it comes to the question of whether an axiom system is consistent. Here, it seems like there is an "objective" answer to whether a certain axiom system will lead to a contradiction (provided we agree upon the logical axioms, inference rules, and proof procedure) – there either is a finite derivation that leads to a contradiction, or there is not. What I find puzzling, however, is that from a purely formal viewpoint, the arithmetic statement $\operatorname{con}(\mathsf{PA})$ seems no different to other arithmetic statements that are independent of $\mathsf{PA}$. So, following the line of reasoning expressed in the first paragraph, it seems that I should believe that $\mathsf{con}(\mathsf{PA})$ is true in some models of arithmetic, and not in others. This makes me cast doubt about whether $\mathsf{con}(\mathsf{PA})$ "really" says that $\mathsf{PA}$ is consistent – if the latter has an "objective" truth value that doesn't require qualification, then the idea of $\mathsf{con}(\mathsf{PA})$ being true in some models and not in others appears very strange.
So, in light of these objections, what viewpoints do logicians have about the relationship between the assertion that $\mathsf{PA}$ is consistent, and the arithmetic sentence $\mathsf{con}(\mathsf{PA})$?