Almost all instances of independence are instances of your further phenomenon, for the following general reason. Namely, if there is a model of ZFC which believes that "$\sigma$ is independent of ZFC" is true, then there is another model of ZFC which believes that it is false, since any assertion of independence fails in a model of ZFC plus $\neg$Con(ZFC), as there are no independent assertions for an inconsistent theory.

Another way to say this is that if it is consistent with ZFC that your sentence is independent, then this assertion is itself never provable in ZFC, because to prove an instance of independence is to prove consistency, which is impossible by the incompleteness theorem.

One might like to say that *all* instances of independence are instances of your further phenomenon, but unfortunately, we cannot quite prove that without further hypotheses. Indeed, it is consistent with ZFC that there are independent statements, but no instances of Ind(ZFC,Ind(ZFC,$\sigma$)). To see this, consider any model $M$ of ZFC+Con(ZFC)+$\neg$Con(ZFC+Con(ZFC)). Such a model exists by the incompleteness theorem, and in this model we'll have Con(ZFC) and hence all kinds of independent statements, but inside $M$ we will have no models of ZFC+Con(ZFC), and thus $M$ thinks that there are no models of Ind(ZFC,$\sigma$) for any $\sigma$, and hence $M$ thinks Ind(ZFC,Ind(ZFC,$\sigma$)) always fails.