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Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

Given ZFC alone, one can give a negative answer to one of my follow-ups. Specifically, there is no extension of the product measure to all sets with property $H$ that is invariant under permutations of $\omega$, even if we just ask for finite additivity in the extension. For if $\mu$ is a finitely additive extension of $P$ to all sets with $H$, then $\nu$ (defined as above) will be a finitely additive probability measure on $\mathcal P {2^\omega}$ invariant under permutations. A contradiction follows from the fact that $F_2$ is a subgroup of $S_\omega$. More precisely: Let $\langle Q_n \rangle_{n\in\omega}$ be a partition of $\omega$ into countably infinite subsets. Let $\phi_n$ be a bijection of $\omega$ with $Q_n$, and let $X$ be the subset of $2^\omega$ consisting of functions $f:\omega\to \{0,1\}$ such that $f\circ \phi_n \ne f\circ \phi_m$ if $n\ne m$. Then $P(X)=1$ (the probability of a repeat in a countable sequence of uniform i.i.d. random variables is zero).

Now for any permutation $\pi$ of $\omega$, let $\pi^*$ be the permutation such that $\pi^*(\phi_n(m))=\pi^*(\phi_{\pi(n)}(m))$ for all $n$ and $m$. The permutations of the form $\pi^*$ form a subgroup $G$ of $S_\omega$ isomorphic to the full group, and $G$ has no non-trivial fixed points on $X$. Since $S_\omega$ contains a free group of rank 2, there is no finitely additive $G$-invariant probability measure on $X$, which contradicts the fact that $\nu$ is such a measure.

I still don't know: Can one prove a negative answer to the main question in ZFC alone? Can one prove a negative answer to the main question with $\rho_n$-invariance in ZFC alone?

Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

Given ZFC alone, one can give a negative answer to one of my follow-ups. Specifically, there is no extension of the product measure to all sets with property $H$ that is invariant under permutations of $\omega$, even if we just ask for finite additivity in the extension. For if $\mu$ is a finitely additive extension of $P$ to all sets with $H$, then $\nu$ (defined as above) will be a finitely additive probability measure on $\mathcal P {2^\omega}$ invariant under permutations. A contradiction follows from the fact that $F_2$ is a subgroup of $S_\omega$. More precisely: Let $\langle Q_n \rangle_{n\in\omega}$ be a partition of $\omega$ into countably infinite subsets. Let $\phi_n$ be a bijection of $\omega$ with $Q_n$, and let $X$ be the subset of $2^\omega$ consisting of functions $f:\omega\to \{0,1\}$ such that $f\circ \phi_n \ne f\circ \phi_m$ if $n\ne m$. Then $P(X)=1$ (the probability of a repeat in a countable sequence of uniform i.i.d. random variables is zero).

Now for any permutation $\pi$ of $\omega$, let $\pi^*$ be the permutation such that $\pi^*(\phi_n(m))=\pi^*(\phi_{\pi(n)}(m))$ for all $n$ and $m$. Note that $\pi^*[X]=X$. The permutations of the form $\pi^*$ form a subgroup $G$ of $S_\omega$ isomorphic to the full group, and $G$ has no non-trivial fixed points on $X$. Since $S_\omega$ contains a free group of rank 2, there is no finitely additive $G$-invariant probability measure on $X$ (here we use AC), which contradicts the fact that $\nu$ is such a measure.

I still don't know: Can one prove a negative answer to the main question in ZFC alone? Can one prove a negative answer to the main question with $\rho_n$-invariance in ZFC alone?

Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

I don't know if one can prove a negative answer in ZFC.

Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

Given ZFC alone, one can give a negative answer to one of my follow-ups. Specifically, there is no extension of the product measure to all sets with property $H$ that is invariant under permutations of $\omega$, even if we just ask for finite additivity in the extension. For if $\mu$ is a finitely additive extension of $P$ to all sets with $H$, then $\nu$ (defined as above) will be a finitely additive probability measure on $\mathcal P {2^\omega}$ invariant under permutations. A contradiction follows from the fact that $F_2$ is a subgroup of $S_\omega$. More precisely: Let $\langle Q_n \rangle_{n\in\omega}$ be a partition of $\omega$ into countably infinite subsets. Let $\phi_n$ be a bijection of $\omega$ with $Q_n$, and let $X$ be the subset of $2^\omega$ consisting of functions $f:\omega\to \{0,1\}$ such that $f\circ \phi_n \ne f\circ \phi_m$ if $n\ne m$. Then $P(X)=1$ (the probability of a repeat in a countable sequence of uniform i.i.d. random variables is zero).

Now for any permutation $\pi$ of $\omega$, let $\pi^*$ be the permutation such that $\pi^*(\phi_n(m))=\pi^*(\phi_{\pi(n)}(m))$ for all $n$ and $m$. The permutations of the form $\pi^*$ form a subgroup $G$ of $S_\omega$ isomorphic to the full group, and $G$ has no non-trivial fixed points on $X$. Since $S_\omega$ contains a free group of rank 2, there is no finitely additive $G$-invariant probability measure on $X$, which contradicts the fact that $\nu$ is such a measure.

I still don't know: Can one prove a negative answer to the main question in ZFC alone? Can one prove a negative answer to the main question with $\rho_n$-invariance in ZFC alone?

Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

I don't know if one can prove a negative answer in ZFC.

Here is a partial answer.

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

I don't know if one can prove a negative answer in ZFC.