Homotopy between Homeomorphisms Let $X=\mathbb{R}^n$, $\phi_0,\phi_1:X\rightarrow X$ be auto-homeomorphisms.  When can we construct a homotopy
$$
\begin{aligned}
\Phi &:X\times [0,1]\rightarrow X\\
&(x,t)\mapsto \Phi(x,t)
\end{aligned}
$$
such that $\Phi(x,i)=\phi_i(x)$, for $i=0,1$ and for each $t \in [0,1]$, $x\mapsto \Phi(x,t)$ is a homeomorphism?
 A: This is asking about the path components of the space $\operatorname{Homeo}(\mathbb{R}^n , \mathbb{R}^n)$.
Let's do a warm up question which is to compute the path components of $\operatorname{Diffeo}(\mathbb{R}^n , \mathbb{R}^n)$. By translation, the inclusion of the diffeomorphisms fixing the origin is a deformation retraction, so we may assume that our diffeomorphisms fix the origin.
Locally near the origin, the map $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ behaves like its matrix of partial derivatives. By essentially zooming in infinitely far, we give a path from $f$ to $f'$ that depends continuously on $f$. Hence, we have a deformation retraction of $\operatorname{Diffeo}(\mathbb{R}^n , \mathbb{R}^n)$ to $GL_n(\mathbb{R})$, and the latter has two path components which are determined by the degree of the map. See http://people.math.harvard.edu/~kupers/teaching/272x/book.pdf Theorem 9.1.1 for a precise argument.
Much more difficult is the topological case, but from above you might conjecture there are exactly two path components depending on the degree of the map. It relies on a difficult theorem that all degree 1 homeomorphisms of $\mathbb{R}^n$ are stable, meaning they are a composition of homeomorphisms that are the identity on an open subset.
It is easy to prove that a stable homeomorphism is isotopic to the identity, so there are two path components of $\operatorname{Homeo}(\mathbb{R}^n , \mathbb{R}^n)$. For references, I point you to chapter 26 of the above notes by Kupers.
So to directly answer your question, you have such a homotopy, if and only if, both are orientation preserving or both reversing.
