Is every homeomorphism subgroup “representable by homotopies”? I came to the conclusion that I frequently visualized homeomorphisms by embedding the space somewhere and gradually deviating from the identity. This motivates the following:
DEF let $X$ be a topological space, $S\leq \operatorname{Aut}(X)$. $S$ is said to be ($i$-)representable by homotopies if there is an embedding $i$ into some other topological space $Y$ such that for every homeo $h\in S$ the embeddings $i\circ h$ and $i \circ\mathrm{id}_X$ are related by a homotopy $H\colon X\times [0,1]\to Y$ such that every $H(-,t)$ is an embedding as well (it is easy to check that this defines a stronger equivalence relation than simple homotopy).
We can make the following observations:


*

*Sometimes, it suffices to choose $\mathrm{id}$ as the embedding: if we consider $\mathrm U(1)\leq \mathrm{Aut} (S^1)$, surely every of those rotations can be reached by a parametrized rotation starting with a zero angle. However, that doesn't capture all of $\mathrm{Aut}$, since w.g. the antipodal map requires at least an embedding into $\mathbb R^3$.

*Consider the infinite cross $X:=V(xy)\subseteq \mathbb R^2$. Then for homeos which preserve each of the ”arms“, the identity embedding will suffice; once we have rotation-like homeos, we will need the embedding in $\mathbb R^2$, and once we have different things like $(x,y)\mapsto (-x,y)$ we will need the third dimension. Note that the question would be rather trivial hadn't we restricted our homotopies to be embeddings in every step; for otherwise, we could relate every homeo to the constant map $(x,y)\mapsto (0,0)$ due to our space's contractability. By this construction, all of $\mathrm{Aut}(X)$ would be $\mathrm{id}$-realizable already.

*Of course, if $S$ is $i$-representable, and there is an embedding $j\colon Y\to Z$, then $S$ is also $j\circ i$-representable. In this vein, we can define a pre-order structure on representations (i.e., said embeddings representing $S$ by homotopies) of $S$ by saying that $i_1\leq i_2$ if they can be mediated by an embedding $j\colon Y_1\to Y_2$ such that $i_2=j\circ i_1$.
Q given $X$, which $S\leq \mathrm{Aut}(X)$ are representable by homotopies?
Follow-Up:
Q when are those $S$ already represented by $id$?
(I apologize for any typos, unnoticed blunders or trivialities, I'm on my mobile in my bed, and it's kind of late)
 A: $\mathrm{Aut}(X)$ (and thus also any $S \leq \mathrm{Aut}(X)$) is always representable. 
For each $h \in \mathrm{Aut}(X)$, define $Z_h = X \times [0, 1] \times \{h\}$, and let $Z$ be the disjoint union of $X$ together with all $Z_h$. Let $Y$ be the quotient space of $Z$ after attaching both ends of each $Z_h$ to $X$ via the identification $(x, 0, h) \sim x$ and $(x, 1, h) \sim h(x)$. Now, let $i : X \to Y$ be the inclusion (the composition of the inclusion $X \hookrightarrow Z$ with the quotient $q : Z \to Y$), and for a given $h \in \mathrm{Aut}(X)$ consider the map $H_h : X \times [0, 1] \to Y$ given by $H_h(x, t) = (x, t, h)$. Then $H_h(x, 0) = (x, 0, h) \sim x$, while $H_h(x, 1) = (x, 1, h) \sim h(x)$, meaning that $H_h(-, 0) = i \circ \mathrm{id}_X$ and $H_h(-, 1) = i \circ h$, i.e. $H_h$ is a homotopy between $i \circ \mathrm{id}_X$ and $i \circ h$. It's also true that $H_h(-, t)$ is an embedding for any $t$, since the quotient $q : Z \to Y$ is an embedding when restricted to $X \times \{t\} \times \{h\}$.
