Is there a separative forcing notion $\mathbb{P}$ such that:
1) For any $p \in\mathbb{P}, \mathbb{P}/p = \{q \in \mathbb{P}: q \leq p \}$ is not forcing isomorphic to any homogeneous forcing notion,
2) For all $G$, $\mathbb{P}$-generic over $V$, $HOD^{V[G]} \subseteq V$.
Here by homogeneity I mean cone homogeneity, i.e. for any p, q in $\mathbb{P}$, there are $p' \leq p, q' \leq q$ and an isomorphism from $\mathbb{P}/ p' $ onto $\mathbb{P}/q'.$
It is consistent that the answer is positive and it is consistent that the answer is negative.
Claim: There is a generic extension, $V[G]$ by a weakly homogeneous forcing notion in which there is a rigid forcing notion $\mathbb{P}$ such that for every generic filter $H \subseteq \mathbb{P}$, $$HOD^V = HOD^{V[G]} = HOD^{V[G][H]}.$$
Proof: Let $\kappa$ be a regular uncountable cardinal. Recall that $\square(\kappa)$ sequence is a sequence $\mathcal{C} = \langle C_\alpha \mid \alpha < \kappa\rangle$ such that $C_\alpha \subseteq \alpha$ is a club, for every accumulation point $\beta \in \text{acc }C_\alpha$, $C_\beta = C_\alpha \cap \beta$ and there is no club $D \subseteq \kappa$ such that for every $\beta \in \text{acc }D$, $D \cap \beta = C_\beta$.
Let $\mathbb{S}$ be the forcing notion for adding a $\square(\kappa)$ sequence using bounded approximations (or successor ordinal length). Let $\mathcal{C}$ be the generic $\square(\kappa)$ sequence. Let $\mathbb{P}$ be the forcing notion $\mathcal{C}$, where $C_\alpha$ is stronger than $C_\beta$ in the order of $\mathbb{P}$ if $C_\alpha$ is an end extension of $C_\beta$.
$\mathbb{S}$ is weakly homogeneous and $\mathbb{S} \ast \mathbb{P}$ has a dense subset isomorphic to the Cohen forcing $\text{Add}(\kappa, 1)$. Therefore the Boolean completion of $\mathbb{S} \ast \mathbb{P}$ is weakly homogeneous. On the other hand $\mathbb{P}$ is rigid. In fact, if $V[G] \subseteq W$ is a model of $ZFC$ and $H_1, H_2\in W$ are two distinct $V[G]$-generic filters for $\mathbb{P}$ then $W \models \text{cf }\kappa = \omega$.
By density arguments (on $\mathbb{S}$ and $\mathbb{P}$), $D_1 = \bigcup H_1,\ D_2 = \bigcup H_2$ are both threads, namely, for $i = 1, 2$ and every accumulation point $\beta\in \text{acc }D_i$, $D_i\cap \beta = C_\beta$. But if $\text{cf }\kappa$ is uncountable, then $\text{acc }D_1 \cap \text{acc }D_2$ is unbounded at $\kappa$ and therefore $D_1 = D_2$ and thus $H_1 = H_2$. This implies that there is no automorphism for the Boolean completion of $\mathbb{P}$ which is non-trivial on the generic filter, since this automorphism sends the generic filter to a different generic filter. Since $\mathbb{P}$ preserves the regularity of $\kappa$ - this is impossible.
Claim: Assume $V = L$. Then for every complete Boolean algebra $\mathbb{B}$, $\Vdash_{\mathbb{B}} HOD^{V[G]} = L$ iff for densely many $b\in \mathbb{B}$, $\mathbb{B} \restriction b$ is weakly homogeneous.
Proof: Let $$\mathbb{C} = \{ x \in \mathbb{B} \mid \forall \sigma \in \text{Aut}({\mathbb{B}}),\ \sigma(x) = x\}.$$
Where we take the set of all automorphisms that respect also arbitrary $\inf$ and $\sup$.
$\mathbb{C}$ is a complete subalgebra of $\mathbb{B}$. Let $G \subseteq \mathbb{B}$ be $L$-generic and let us claim that the generic filter for $\mathbb{C}$ belongs to $HOD^{V[G]}$.
Indeed, $b \in G\cap \mathbb{C}$ iff for every $L$-generic filter $H\subseteq \mathbb{B}$, $b\in H$. This follows from a theorem of Vopenka and Hajek:
Theorem (Vopenka and Hajek): If $H_1, H_2\subseteq \mathbb{B}$ are $V$-generic and $V[H_1] = V[H_2]$ then there is an automotphism $\sigma\in \text{Aut}(\mathbb{B})$ such that $\sigma '' H_1 = H_2$.
Thus, we conclude that $\Vdash \dot{G} \cap \mathbb{C}\in L$, and in particular, in $\mathbb{C}$ there is a dense set of atoms. Otherwise, there is $c\in \mathbb{C}$ such that there are no atoms below $c$ in $\mathbb{C}$. Therefore, $G \cap \mathbb{C} \notin L$ for every generic filter $G$ that contains $c$, since it is a generic filter for non-atomic forcing. Let $a$ be an atom of $\mathbb{C}$. The forcing $\mathbb{B} \restriction a$ is weakly homogeneous, as for every $b \leq a$, the orbit of $b$, $\{\sigma(b) \mid \sigma \in \text{Aut}(\mathbb{B})\}$ is below $a$, and therefore $\bigvee_{\sigma \in \text{Aut}(\mathbb{B})} \sigma(b) = a$. Therefore, for every $c \leq a$, there is $\sigma$ such that $\sigma(b) \wedge c \neq 0$.
