Homogeneous Namba-like forcing Let $\kappa \ge \aleph_3$ be a regular cardinal that is countably closed ($\alpha^\omega < \kappa$ for every $\alpha < \kappa$.)  I'm mostly interested in the case that $\kappa$ is strongly inaccessible.  Can there be a homogeneous notion of forcing that makes $\text{cof}(\kappa^{+V}) < \kappa$ without adding any bounded subsets of $\kappa$?
If there is a Woodin cardinal above $\kappa$ then the stationary tower forcing could do this except that it is (probably) not homogeneous.
If there is a forcing notion as desired then I believe the results of the paper "Stacking mice" would give a non-domestic mouse, so some large cardinals would be required to show that such a forcing exists.  Can we get one from, e.g. a supercompact cardinal?
 A: I think that the strongly compact Prikry forcing, for $\kappa^+$ strongly compact cardinal $\kappa$ - that forces $\text{cf }\kappa = \text{cf }(\kappa^+)^V = \omega$ without adding bounded subsets to $\kappa$, is homogeneous, but I couldn't prove it or find a reference.
Instead, I'll show something weaker that still implies that the truth value of statements of the form $\phi(a)$ in $V[G]$, where $a\in V$ doesn't depend on the generic $G$: For every $p,q\in \mathbb{P}$, we will find $p^\prime \leq p,\,q^\prime \leq q$ and an automorphism between $\mathbb{P}\restriction p^\prime = \{r \in \mathbb{P} | r \leq p^\prime\}$ and $\mathbb{P}\restriction q^\prime$.
Let $\mathbb{P}$ be the strongly compact Prikry forcing for changing both $\text{cf }\kappa$ and $\text{cf }\kappa^+$ to $\omega$, without adding bounded subsets to $\kappa$. 
Recall that a condition in $\mathbb{P}$ is a tree of finite increasing sequences in $P_\kappa (\kappa^+ )$, with finite trunk and above it every element has $U$-many successors ($U$ is fine $\kappa$-complete ultrafilter over $P_\kappa (\kappa^+)$). For the exact definitions and basic properties see section 1.4 in Gitik's chapter in the Handbook. 
Let $t$ be the trunk of $p$ and $s$ the trunk of $q$. By narrowing the trees of $p$ and $q$, if necessary, we may assume that for every $r\in p$ above $t$, $(r\setminus t)\cup s \in q$. For every $a\leq p$ define $\pi (a) = \{(r\setminus t)\cup s | r\in a\}$ - this is the required automorphism.
