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Yes. We can always extend a function $f:\{0,1\}^n\rightarrow\{0,1\}^n$ to a homeomorphism $f:[0,1]^n\rightarrow[0,1]^n$. I am sure that this can be made computable, but I have not yet thought about an explicit formula for an extension.

Proposition: Suppose that $X$ is a connected regular topological space where $X\setminus\{x_0\}$ is connected for each $x_0\in X$. Suppose furthermore that whenever $x_0\in U$ and $U$ is an open set, there is an open set $V$ with $x_0\in V\subseteq\overline{V}\subseteq U$ such that if $y_0\in V$, then there is a homeomorphism $f:X\rightarrow X$ such that $f(x_0)=y_0$ but where $f$ is the identity function outside of $\overline{V}$. Then whenever $x_1,\dots,x_n\in X$ are distinct and $y_1,\dots,y_n\in X$ are distinct, there is a homeomorphism $f:X\rightarrow X$ where $f(x_j)=y_j$ for $1\leq j\leq n$.

Proof: If $n$ is a natural number, then let $X^{\langle n\rangle}$ be the subspace of $X^n$ consisting of all tuples $(x_1,\dots,x_n)$ of distinct elements. Then $X^{\langle n\rangle}$ is connected for all $n\geq 1$. Now, given $(x_1,\dots,x_n)\in X^{\langle n\rangle}$, let $O_{x_1,\dots,x_n}$ be the collection of all tuples $(y_1,\dots,y_n)\in O_{x_1,\dots,x_n}$ where there is some homeomorphism $f:X\rightarrow X$ where $y_j=f(x_j)$ for $1\leq j\leq n$. I claim that $O_{x_1,\dots,x_n}$ is an open subset of $X^{\langle n\rangle}$. Suppose that $(y_1,\dots,y_n)\in O_{x_1,\dots,x_n}$. Then let $U_1,\dots,U_n$ be disjoint open sets with $y_j\in U_j$ for $1\leq j\leq n$. there are open sets $V_1,\dots,V_n$ with $y_j\in V_j\subseteq\overline{V_j}\subseteq U_j$ and homeomorphisms $f_1,\dots,f_n:X\rightarrow X$ where $f_j(y_j)=z_j$ for $1\leq j\leq n$ but where $f_j$ is the identity function outside $\overline{V_j}$. In this case, we can glue these homeomorphisms and obtain a homeomorphism $f:X\rightarrow X$ with $f|_{\overline{V_j}}=f_j|_{\overline{V_j}}$ and where $f$ is the identity function outside $\overline{V_1}\cup\dots\cup\overline{V_n}$. In this case, $f(y_j)=z_j$ for $1\leq j\leq n$. Now, if $g:X\rightarrow X$ is a homeomorphism with $g(x_j)=y_j$ for $1\leq j\leq n$. Then $(f\circ g)(x_j)=z_j$ for $1\leq j\leq n$. We may therefore conclude that $(z_1,\dots,z_n)\in O_{x_1,\dots,x_n}$. We can therefore conclude that $O_{x_1,\dots,x_n}$ contains an open neighborhood of $(y_1,\dots,y_n)$, so $O_{x_1,\dots,x_n}$ is an open subset of $X^{\langle x\rangle}$. However, $(O_{x_1,\dots,x_n})_{(x_1,\dots,x_n)\in X^{\langle n\rangle}}$ is a partition of the connected space $X^{\langle n\rangle}$ into open sets. This is only possible if $O_{x_1,\dots,x_n}=X^{\langle n\rangle}$. $\square$

In particular, if $M$ is a manifold without boundary of dimension at least $2$, then whenever $(x_1,\dots,x_n),(y_1,\dots,y_n)\in M^{\langle n\rangle}$, there is a homeomorphism $f:M\rightarrow M$ with $f(x_j)=y_j$ for $1\leq j\leq n$. Since $\partial [0,1]^n$ is homeomorphic to $S^{n-1}$, every $f:\{0,1\}^n\rightarrow\{0,1\}^n$ can be extended to a homeomorphism $g:\partial [0,1]^n\rightarrow \partial [0,1]^n$ (when $n\geq 3$) which can then be extended to a homeomorphism $g:[0,1]^n\rightarrow[0,1]^n$.