Codimension zero embeddings and diffeomorphism groups Let $V$ be a smooth manifold obtained by attaching the ``open collar'' $[0,1)\times \partial N$ to a compact smooth manifold $N$ along the boundary. Let $\mathrm{Emb}(N, V)$ be the space of embeddings of $N$ into $V$ isotopic to the inclusion. The identity component $\mathrm{Diff\,} N$ of the diffeomorphism group of $N$ embeds into $\mathrm{Emb}(N, V)$ by precomposing a diffeomorphism with the inclusion.
Question: Under what conditions the embedding $\mathrm{Diff\,} N\to\mathrm{Emb}(N, V)$ is not a homotopy equivalence?
It is relevant to consider the $\mathrm{Diff\,} V$-action on $\mathrm{Emb}(N, V)$ by postcomposing. The orbit map $\mathrm{Diff\,} V\to\mathrm{Emb}(N, V)$ is a homotopy equivalence: its homotopy fiber consists of diffeomorphisms that are identity on $N$, and the group of such diffeomorphisms is contractible by the Alexander trick towards infinity. 
The case when $V=\mathbb R^n$ and $N=D^n$ is fairly well 
understood: both $\mathrm{Diff\,} \mathbb R^n$ and $\mathrm{Diff\,}\mathbb D^n$ deformation retract onto $O(n)$ [Edit: as Allen Hatcher explains in comments $\mathrm{Diff\,}\mathbb D^n$ does not deformation retract onto $O(n)$]. On the other hand, $\mathrm{Diff\,}(\mathbb D^n, rel\,\partial D^n)$ is homotopically more complicated when $n$ large (computing it involves pseudoisotopy and Waldhausen K-theory).
Naively I hope the case of a disk be used to produce elements of $\pi_i(\mathrm{Diff\,} N)$ that die in $\pi_i(\mathrm{Emb}(N,V))\cong\pi_i(\mathrm{Diff\,} V)$, e.g. by considering boundary connected sum of $N$ with $D^n$, and extending diffeomorphisms of $D^n$ that are identity on $\partial D^n$ to $N$ by the identity. Has this ever been done?  
I am even more interested in ways to produce elements of $\pi_i(\mathrm{Emb}(N,V))$ that do not come from $\pi_i(\mathrm{Diff\,} N)$.
 A: [The following is an elaboration of my comment above, in response to Igor Belegradek's inquiries. Due to the typographical limitations of comments, I am posting it as an answer.]$\DeclareMathOperator{\Diff}{Diff}$$\DeclareMathOperator{\Emb}{Emb}$$\DeclareMathOperator{\interior}{int}$$\newcommand{\To}{\longrightarrow}$
As is essentially pointed out in this answer by Tom Goodwillie, the homotopy fibre of the map $\Diff(⁡N) \to \Emb⁡(N,V)$ is homotopy equivalent to the pseudo-isotopy space of $\partial N$, i.e. the space of diffeomorphisms of $\partial N \times I$ which fix $\partial N \times\{0\}$ pointwise.
Now I will briefly explain why that conclusion follows from Tom Goodwillie's answer. As in Tom's answer, let $N_0 \subset N$ be the complement of an open collar of $\partial N$ in $N$. More precisely, if we are given some embedding $\partial N \times [0,1) \to N$ which is the inclusion of the boundary when restricted to $\partial N \times \{0\}$, let $N_0$ be $N$ minus the image of $N\times[0,\frac12)$. Then we have the map described by Tom in his answer,
$$ r = k^\ast : \Diff(N) \To \Emb(N_0,\interior N) $$
which is given by restriction along $k:N_0 \to N$.
On the other hand, consider, as in the question, a manifold $V$ obtained from $N$ by gluing an open collar along the boundary of $N$. Then we also have the map defined in the question
$$ i = l_\ast : \Diff(N) \To \Emb(N,V) $$
given by composing (on the left) with the inclusion $l:N \to V$.
We only need to consider the following commutative square diagram
$$
\begin{matrix}
\Diff(N) & \overset{i}{\To} & \Emb(N,V) \\
\Big\downarrow\rlap{r} &  & \Big\downarrow\rlap{s} \\
\Emb(N_0,\interior N) & \underset{j}{\To} & \Emb(N_0,V)
\end{matrix}
$$
where $s = k^\ast : \Emb(N,V) \to \Emb(N_0,V)$ is the restriction map, and $j = m_\ast : \Emb(N_0,\interior N) \to \Emb(N_0,V)$ is given by composition (on the left) with the inclusion $m:\interior N \to V$.
Finally, observe that both $s$ and $j$ are homotopy equivalences (one needs to use the collars to show this). Therefore, the homotopy fibre of $r$ is homotopy equivalent to that of $i$. On the other hand, the map $r$ is a Hurewicz fibration, as follows from a version of the parametrized isotopy extension theorem. Hence, the homotopy fibre of $r$ is equivalent to its fibre. The desired conclusion is reached because the fibre of $r$ is canonically homeomorphic to the pseudo-isotopy space of $\partial N$.
