There is a locally-trivial fibre bundle

$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$

here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$. $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$.

First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.

So now you have a fibration:

$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$

where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.

$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there. In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(X_e)$ allows on the spherical normal bundle to $L$. There's probably a decent answer to that which doesn't take too much work but the above is a start.

There is a locally-trivial fibre bundle

$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$

here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$. $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$.

First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.

So now you have a fibration:

$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$

where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.

$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there. In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(X_e)$ allows on the spherical normal bundle to $L$. There's probably a decent answer to that which doesn't take too much work but the above is a start.

edit: by Cerf's pseudoisotopy theorem, the kernel of the map $\pi_0 Diff(S^{n+2}, L) \to \pi_0 Diff(X_e)$ contains the exotic sphere "part" of $\pi_0 Diff(S^{n+2}, L)$.

There is a locally-trivial fibre bundle

$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$

here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ the the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$.

First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.

So now you have a bundle where the induced maps on homotopy groups are a short-exact sequence:

$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$

in particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.

$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions as the spherical normal bundle to $L$ is particularly symmetric in low dimensions. In general it is a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(S^{n+2}, L)$ allows on the spherical normal bundle to $L$. Anyhow, there's probably a decent answer to that but the above is a start.

There is a locally-trivial fibre bundle

$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$

here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$. $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$.

First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.

So now you have a fibration:

$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$

where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.

$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there. In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(X_e)$ allows on the spherical normal bundle to $L$. There's probably a decent answer to that which doesn't take too much work but the above is a start.