Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).

Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)

Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$?

May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...) to guarantee uniqueness ?

Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. Is it at least true that two quantizations living in the same orbit of that group give isomorphic quantum algebras ?

Question in formally precise form Consider a Poisson algebra. Choose two different formality isomorphisms (e.g. with different propagators or associators).

Define two star-products $\star$ and $\star'$ with the help of these two formality isomorphisms.

Question Are the algebras defined by these two star-products isomorphic ? More strongly - are these star products "equivalent" ? (See definition of equivalence in Stefan Waldmann's answer below or in Kontsevich paper).

Some comments.

If our manifold is R^2n we canonical Poisson bracket { p_i q_j } = delta_ij, then undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_{ij}$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird, it would be non-obvious that we get isomorphism with Heisenberg algebra. And moreover it might depend on category we are working with (polynomial or smooth functions).

More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).

Concerning choices of different coordinates in classical algebra A - it already states in Kontsevich paper that obtained algebras will be isomorphic. More strongly star-products will be "equivalent".
See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators

I had discussed this question with some experts some years ago, but there was no clear answer.

The motivation to ask partly comes from MO-discussions here: Quantization of a classical system (e.g. the case of a billiard)

Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).

Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)

Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$?

May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...) to guarantee uniqueness ?

Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. Is it at least true that two quantizations living in the same orbit of that group give isomorphic quantum algebras ?

Question in formally precise form Consider a Poisson algebra. Choose two different formality isomorphisms (e.g. with different propagators or associators).

Define two star-products $\star$ and $\star'$ with the help of these two formality isomorphisms.

Question Are the algebras defined by these two star-products isomorphic ? More strongly - are these star products "equivalent" ? (See definition of equivalence in Stefan Waldmann's answer below or in Kontsevich paper).

Some comments.

If our manifold is R^2n we canonical Poisson bracket { p_i q_j } = delta_ij, then undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_{ij}$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird, it would be non-obvious that we get isomorphism with Heisenberg algebra. And moreover it might depend on category we are working with (polynomial or smooth functions).

More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).

Concerning choices of different coordinates in classical algebra A - it already states in Kontsevich paper that obtained algebras will be isomorphic. More strongly star-products will be "equivalent".
See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators

I had discussed this question with some experts some years ago, but there was no clear answer.

The motivation to ask partly comes from MO-discussions here: Quantization of a classical system (e.g. the case of a billiard)

Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).

Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)

Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$?

May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...) to guarantee uniqueness ?

Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. Is it at least true that two quantizations living in the same orbit of that group give isomorphic quantum algebras ?

Some comments.

If our manifold is R^2n we canonical Poisson bracket { p_i q_j } = delta_ij, then undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_ij$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird, it would be non-obvious that we get isomorphism with Heisenberg algebra. And moreover it might depend on category we are working with (polynomial or smooth functions).

More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).

Concerning choices of different coordinates in classical algebra A - it already states in Kontsevich paper that obtained algebras will be isomorphic. See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators

I had discussed this question with some experts some years ago, but there was no clear answer.

The motivation to ask partly comes from MO-discussions here: Quantization of a classical system (e.g. the case of a billiard)

Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).

Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)

Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$?

May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...) to guarantee uniqueness ?

Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. Is it at least true that two quantizations living in the same orbit of that group give isomorphic quantum algebras ?

Question in formally precise form Consider a Poisson algebra. Choose two different formality isomorphisms (e.g. with different propagators or associators).

Define two star-products $\star$ and $\star'$ with the help of these two formality isomorphisms.

Question Are the algebras defined by these two star-products isomorphic ? More strongly - are these star products "equivalent" ? (See definition of equivalence in Stefan Waldmann's answer below or in Kontsevich paper).

Some comments.

If our manifold is R^2n we canonical Poisson bracket { p_i q_j } = delta_ij, then undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_{ij}$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird, it would be non-obvious that we get isomorphism with Heisenberg algebra. And moreover it might depend on category we are working with (polynomial or smooth functions).

More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).

Concerning choices of different coordinates in classical algebra A - it already states in Kontsevich paper that obtained algebras will be isomorphic. More strongly star-products will be "equivalent". 
See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators

I had discussed this question with some experts some years ago, but there was no clear answer.

The motivation to ask partly comes from MO-discussions here: Quantization of a classical system (e.g. the case of a billiard)