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In the previous paper [K. S. Lau, J. J. Luo and H. Rao, Topological structure of fractal squares, Math. Proc. Camb. Phil. Soc.155 (2013) 73–86], Lau, Luo and Rao completely classified the topological structure of so called fractal square$F$ defined by $F=(F+{𝒟})/n$, where ${𝒟}⊊{\{0,1,\dots ,n-1\}}^2,n\ge 2$. In this paper, we further provide simple criteria for the $F$ to be totally disconnected, then we discuss the Lipschitz classification of $F$ in the case $n=3$, which is an attempt to consider non-totally disconnected sets.

A fractal square is a nonempty compact set in the plane satisfying $F=(F+D)/n$, where $n>1$ is an integer and $D\subset {\{0,1,2,\dots ,n-1\}}^2$ is nonempty. We give the topological classification of fractal squares with $n=3$ and $\text{Card}(D)=6$.

Given an integer $n\ge 2$ and a digit set ${𝒟}\subset {\{0,1,\dots ,n-1\}}^2$, there is a self-similar set $K\subset {ℝ}^2$ satisfying the set equation $K=(K+{𝒟})/n$. This set $K$ is called a fractal square. By studying the line segments contained in $K$, we give a lower estimate of the topological Hausdorff dimension of fractal squares. Moreover, we compute the topological Hausdorff dimension of fractal squares whose nontrivial connected components are parallel line segments, and introduce the Latin fractal squares to investigate the question when the topological Hausdorff dimension of a fractal square coincides with its Hausdorff dimension.

We consider a family of fractal squares, denoted as ${{ℱ}}_{3,7}$. Each of them satisfies the set equation $K=\frac{1}{3}(K+{𝒟})$ for some ${𝒟}\subset {\{0,1,2\}}^2$ with $\#{𝒟}=7$. It is known that two of these fractal squares are Lipschitz equivalent if and only if they are isometrically equivalent. The aim of our study is to improve this by replacing Lipschitz equivalence with topological equivalence. To this end, we shall investigate the group $G_{\mathrm{\text{aut}}}(K)$ of all homeomorphisms of a fractal square $K\in {{ℱ}}_{3,7}$ that has a cut point and show that $\#G_{\mathrm{\text{aut}}}(K)=2$ or $8$.