Let $M$ and $N$ be smooth manifolds. Consider an isotopy of $M$ inside $N$. This means that we have a level preserving embedding $J\colon M\times [0,1] \to N \times [0,1]$. Put $J(x,t)=(\phi\_t(x),t)$. Hence $\phi\_t\colon M \to N$ is an embedding for each $t\in [0,1]$.


If $M$ is compact and $N$ has no boundary, then the classical isotopy extension theorem states that there exists a diffeotopy (aka ambient isotopy) $H\colon M \times [0,1] \to M$ (i.e. a 1-parameter family of diffeomorphisms), extending $J$, i.e. such that 


a) $\phi\_t(x)=H(\phi\_0(x),t)$, for $x \in M$ and $t \in [0,1]$, and 


b) $H(y,0)=y$, for each $y \in N$.


Question: how unique is such ambient isotopy $H$? Can any two ambient isotopies $H$ and $H'$ extending $J$ always be connected by some form of second order ambient isotopy?


(The main example I am thinking of is that of isotopies of a disjoint union of 1-spheres inside the 3-sphere.) 



Isotopy extension theorem: how non-unique is ambient isotopy
If I interpret the question correctly then the answer is "yes". You seem to be asking whether, if $H'$ is an isotopy satisfying the same conditions as $H$, there must be a one-parameter family of such isotopies joining $H$ to $H'$. I claim that the space of all such isotopies $H'$ is not only path-connected but contractible.


Write $H'(y,t)=H(K(y,t))$. This defines an isotopy $K:N\times I\to N\times I$, such that $K(y,0)=(y,0)$ for all $y$, and such that $K(\phi\_0(x),t)=(\phi\_0(x),t)$ for all $x$ and $t$. So $K$ is a one-parameter family, starting at $1\_N$, in the space of all diffeomorphisms of $N$ that fix $\phi\_0(M)$ pointwise. The space of all such $K$ is contractible because more generally for any space $X$ and any point $p\in X$ the space of all paths in $X$ starting at $p$ is contractible.


