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The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$

The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a-(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$-dimensional vector space, if I'm not mistaken, so the question here is: are any of them actually isometries? Again if I'm not mistaken, the answer is no. But these things you should check yourself, because I really didn't.

The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$

The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a-(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$-dimensional vector space, so the question here is: are any of them actually isometries? Again the answer is no (it is interesting to check it). So you have found all conformal vector fields as either Killing or of the form $X_a(x)$: $$\mathfrak{conf}(\mathbb{S}^d) = \mathfrak{isom}(\mathbb{S}^d) \oplus \{X_a | a\in \mathbb{R}^{d+1}\} $$

The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$

The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a-(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$-dimensional vector space, if I'm not mistaken, so the question here is: are any of them actually isometries? Again if I'm not mistaken, the answer is no.

The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$

The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a-(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$-dimensional vector space, if I'm not mistaken, so the question here is: are any of them actually isometries? Again if I'm not mistaken, the answer is no. But these things you should check yourself, because I really didn't.

The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}-1$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

The above isomorphism should in particular tell you how to construct explicitly more of such infinitesimal conformal symmetries.

The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2}| -x_0^2+x_1^2+\ldots+x_{d+1}^2=-1\}$ and $B^{d+1}$ is the $(d+1)$-dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2|dx|}{1-|x|^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.

All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).

Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$

The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a-(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$-dimensional vector space, if I'm not mistaken, so the question here is: are any of them actually isometries? Again if I'm not mistaken, the answer is no.