Let $I = [0,1]$. For this post start the ordinals at $1$ and consider this a limit ordinal. Consider an ordinal $\lambda$ and let $\lambda_{*}$ be the greatest limit ordinal less than or equal to $\lambda$; so $\lambda = \lambda_{*} + n$ for some nonnegative integer $n$. Consider $I^{\lambda}$ as a topological space in the dictionary order. Let $\mathbb{0}$ denote a sequence of $\lambda_{*}$ zeroes and define $\mathbb{1}$ analagously with ones. (So $\mathbb{0} \times I^n$ denotes the subset of $I^{\lambda}$ whose elements begin with $\lambda_{*}$ zeroes and vary arbitrarily in the last $n$ coordinates.)
The title is actually incorrect when $\lambda$ is not a limit ordinal but I wasn't sure how to word it descriptively and succinctly (see first comment). I'm really asking the following. Is $I^{\lambda} - (\mathbb{0} \times I^n \cup \mathbb{1} \times I^n)$ homogeneous? Is it 2-transitive in the case of a limit ordinal?
I only have intuition for the finite cases and for the limit ordinal cases but from these this seems reasonable.
My idea is to argue something like the space $\mathbb{Z} \times (I^{\lambda} - \mathbb{1} \times I^n)$ under the dictionary order is homeomorphic to this space and has a compatible abelian group operation (the obvious one using base number arithmetic), but don't know how to make any of this precise.
Edit: Niels points out that successor ordinals and ordinals with uncountable cofinality will be counterexamples. I should have caught the first example but wouldn't have considered the second and so this question may have been overambitious/naive. In order to give some direction I'd like to narrow the question to the following. Is the space $I^{\omega}$ minus its end-points homogeneous? I suspect it is not because sequences converging to $(\frac{1}{2}, \frac{1}{2}, \ldots)$ from below appear to have a characteristic difference from those sequences converging to $(1,0,0,\ldots{})$ from below but I'm not sure how to make this precise and want to learn.