Sharply $k$-transitive actions on spheres

First, I'm always thinking of $k\geq 1$ and $n\geq 1$. With that in mind, the result is the following:

A topological group $G$ acts sharply $k$-transitively on $S^n$ iff $(n,k) \in \{(1,1), (1,2), (1,3), (2,3), (3,1), (3,2)\}.$

According to your link above, you cannot have a sharply $k$-transitive action for any $k\geq 4$, so we only need to understand what happens for small $k$.

To that end, recall that for a topological space $X$, the configuration space of ordered $k$-tuples in $X$ is the set $X^k\setminus \Delta$, where $\Delta = \{(x_1,...,x_k)\in X^k: x_i\neq x_j\text{ for any } i\neq j\}$. I'll write $C^k X$ for this space. Note that if $X$ is a smooth manifold, then $C^k X\subseteq X^k$ is an open subset, so is naturally a smooth manifold.

Proposition: A topological group $G$ acts sharply $k$-transitively on $S^n$ iff $C^k S^n$ has the structure of a Lie group.

Proof: A $G$-action on $S^n$ gives rise to a $G$-action on $C^k S^n$ by the formula $g\ast(x_1,...,x_k) = (g x_1 , ... g x_k)$. Then a sharply $k$-transitive action of $G$ on $S^n$ is obviously the same as a simply transitive action of $G$ on $C^k S^n$.

Now, if $C^k S^n$ is a Lie group, then taking $G = C^k S^n$, $G$ acts on itself simply transitively by left multiplication, so acts on $S^n$ sharply $k$-transitively.

Conversely, if the $G$-action is sharply $k$-transtiive, then fixing a point $p\in C^k S^n$, the map $G\rightarrow C^k S^n$ given by $g\mapsto g\ast p$ is a homeomorphism. Thus, $C^k S^n$ has a topological group structure. But $C^k S^n$ is locally Euclidean, so using the solution of Hilbert's 5th problem (due to Gleason, Montgomery, and Zippen), $C^k S^n$ is a Lie group. $\square$.

Now, if $k=1$, then $C^1 S^n = S^n$, and, as you said, this is only a Lie group when $n=1,3$.

For $k=2$, we have the following proposition.

Proposition: If $k=2$, then $n = 1$ or $n=3$. Conversely, there are sharply $2$-transitive actions of $\mathbb{R}\times S^1$ on $S^1$, and of $\mathbb{R}^3\times S^3$ on $S^3$.

Proof: The projection map onto the first factor $X^2\rightarrow X$ induces a map $C^2 X\rightarrow C^1 X$ which is a fiber bundle with fiber $X\setminus\{p\}$. When $X = S^n$, this shows that $C^2 S^n$ is an $\mathbb{R}^{n}$-bundle over $S^n$. In particular, $C^2 S^n$ deformation retracts onto $S^n$, and so, $H^n(C^2 S^n;\mathbb{Q})$ is the first non-trivial reduced cohomology group.

Now, every connected non-compact Lie group deformation retracts onto its maximal compact subgroup, and every compact Lie group of positive dimension has $H^1$ or $H^3$ non-trivial. In particular, since $H^n(C^2 S^n; \mathbb{Q})\neq 0$, $n=1$ or $n=3$.

For the converse, note that if $H$ is a Lie group, then $C^2 H$ is diffeomorphic to $(H\setminus\{e\}) \times H.$ Indeed, the map $(H\setminus\{e\})\times H\rightarrow C^2 H$ given by $(h_1, h_2)\mapsto (h_1 h_2, h_2)$ is a diffeomorphism. Thus, $C^2 S^1 \cong \mathbb{R}\times S^1$ and $C^2S^3\cong \mathbb{R}^3\times S^3$. Since both $\mathbb{R}\times S^1$ and $\mathbb{R}^3\times S^3$ are Lie groups, these Lie groups act sharply $2$-transitively on $S^1$ and $S^3$, by the first proposition. $\square$

Let's move onto the case $k=3$.

Proposition: The case $k=3$ arises iff $n\leq 2$.

First, assume $k=3$. The third projection map $S^n\times S^n\times S^n\rightarrow S^n$ gives rise to a map $C^3 S^n\rightarrow S^n$ which is a fiber bundle with fiber $C^2(S^n\setminus \{pt\})$. Since $S^n\setminus \{pt\}$ is diffeomorphic to $\mathbb{R}^n$, which is a Lie group, $C^2 (S^n\setminus \{pt\})$ is diffeomorphic to $(\mathbb{R}^n\setminus\{0\})\times \mathbb{R}^n$.

As $\mathbb{R}^n\setminus\{0\}$ deformation retracts to $S^{n-1}$, it follows that, up to homotopy, $C^3 S^n$ is a bundle over $S^n$ with fiber $S^{n-1}$. So, $H^{n-1}(C^3 S^n;\mathbb{Q})$ is the lowest non-trivial reduced rational cohomology group of $C^3 S^n$. By the proof of the second proposition, this implies that $n-1 \in \{0,1,3\}$, so $n\in \{1,2,4\}$. But, if $n=4$, then $H^4(C^3 S^4;\mathbb{Q})\cong \mathbb{Q}$, which contradicts that every Lie group has the rational cohomology ring of a product of odd dimensional spheres. Thus, $n=4$ cannot occur, so $n\in\{1,2\}$.

Conversely, we need to find examples of strictly $3$-transitive actions on $S^n$ for $n\neq 2$. You've already handled the case where $n=2$.

When $n=1$, $C^3 S^1 \cong S^1\times C^2 \mathbb{R}\cong S^1\times (\mathbb{R}\setminus\{0\})\times \mathbb{R}$. Since this is a Lie group (using multiplication on $\mathbb{R}\setminus \{0\})$, we are done by the first Proposition. $\square$

The purpose of this post is to classify manifolds (not just spheres) which are sharply $3$-transitive. The main result is that the examples in my other answer are the only ones.

Suppose $M$ is a closed manifold which is sharply $3$-transitive. Then $M$ is diffeomorphic to $S^1$ or $S^2$.

Proof: Suppose $G$ acts on $M$ sharply $3$-transitively. From the classifcation of closed $1$-manifolds, we may assume $\dim M\geq 2$. Pick $x\in M$. Then the isotropy group $G_x$ must act transitively on $M\setminus \{x\}$. Pick $y\in M\setminus\{y\}$. Then the isotropy group $(G_x)_y$ must act simply transitively on $N:=M\setminus \{x,y\}$.

This implies that $N$ is diffeomorphic to the Lie group $(G_x)_y$. Note also that $N$ is non-compact and it is connected since $\dim N =\dim M \geq 2$. A connected Lie group is always diffeomorphic to a product $\mathbb{R}^k\times K$ where $K$ is a compact Lie group, so $N\cong \mathbb{R}^k\times K$.

Note that $N$ has a two-point compactification $(M)$, so by this MO answer, $N$ must have at least two ends. This implies $k=1$ because for $k\geq 2$, $N$ has just one end.

Now, the two point compactification of $N$ obtained by adding a point at each end (that is, $M$), is nothing but the suspension of $K$, $\Sigma K$. However, according to this MSE question if $\Sigma K$ is a manifold, then $K$ must have been a sphere. (Technically, the linked answer assumed $\dim K \geq 3$, but if $\dim = 1,2$, then a closed homology sphere must be homeomorphic to a sphere by the low dimensional classification of manifolds.)

Thus, $K$ is a compact Lie group and a sphere. It follows that $K = S^3$ or $K= S^1$. If $K= S^1$, then $M$ is the suspension of $S^1$, so is $S^2$. It $K = S^3$, then $M\cong S^4$ is the suspension of $S^3$. But this was ruled out in the previous answer. $\square$

The result for $k=2$ is the following.

Suppose $M$ is a closed manifold which is sharply $2$-transitive. Then $M$ is diffeomorphic to $S^1$ or $S^3$.

Proof: Suppose $G$ acts on $M$ in a sharply $2$-transitive manner. Picking $x\in M$, we deduce that the isotropy group $G_x$ acts simply transitively on $N:=M\setminus\{x\}$. In particular, $N$ is a non-compact connected manifold, which is diffeomorphic to $G$. Just as above, it follows that $N\cong \mathbb{R}^\ell\times K$ for some compact Lie group $K$.

Now, $M$ is the one point compactification of $N$. So, we may express $M = S^k\times K/(\{\infty\}\times K$. Using the long exact sequence of the pair $(\{\infty\}\times K, M)$, it follows easily that $H^{\ell + s}(K)\cong \overline{H}^s(M)$ for all $s > 0$. (This is just a simple manifestation of the Thom isomorphism applied to the bundle $\mathbb{R}^\ell\rightarrow N\rightarrow K$.)

On the other hand, computing the local homology of the pair $(M,N)$ yields $H^\ast(M,N)\cong \mathbb{Z}$ when $\ast = \dim M$, and $0$ otherwise (since $M$ is a manifold). From the long exact sequence of cohomology of a pair, together with the fact that $N$ deformation retracts to $K$, we see that $H^s(M)\rightarrow H^s(K)$ is an isomorphism for any $0 < s < \dim M$. Combining this with the result in the previous paragraph, we see $H^s(K) \cong H^s(M)\cong H^{s+\ell}(K)$ is valid for any $0<s< \dim M$.

If $\dim K\neq 0$, set $s = \dim K$. Then $H^{\dim K}(K)\neq 0$ because $K$ is orientable (since it's a Lie group), while $H^{s + \ell}(K) = 0$ because $\ell > 1$ (since $\mathbb{R}^\ell\times K$) is non-compact.

Thus, we have a contradiction unless $\dim K = 0$. Now it follows that $N$ is diffeomorphic to $\mathbb{R}^{\ell}$, so $M$, being the one point compactification of $N$, is $S^{\ell}$. Having reduced to the case where $M$ is a sphere, the previous answer now gives the result. $\square$