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Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n+1),\quad \mathrm{SU}(n+1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ and uses the same formula as before, but now, one allows $J$ to be scalar multiplication by any unit imaginary quaternion. After conjugating by an element of $\mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$, all of these actions are isometric, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure in the third case.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.) When $U=\mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)$, one could have $G = \mathrm{SL}(\tfrac{n+1}2,\mathbb{H}\bigr)$, and so on.

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n+1),\quad \mathrm{SU}(n+1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ and uses the same formula as before, but now, one allows $J$ to be scalar multiplication by any unit imaginary quaternion. After conjugating by an element of $\mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$, all of these actions are isometric, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure in the third case.

Thus, there are really only three cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the first three groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold.

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.

Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.

By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.

By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n+1),\quad \mathrm{SU}(n+1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ and uses the same formula as before, but now, one allows $J$ to be scalar multiplication by any unit imaginary quaternion. After conjugating by an element of $\mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$, all of these actions are isometric, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure in the third case.

Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.

This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.

For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.) When $U=\mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)$, one could have $G = \mathrm{SL}(\tfrac{n+1}2,\mathbb{H}\bigr)$, and so on.