Is $\mathrm{Cantor \, set} \times [0,1]$ self-similar? Consider the Cantor set $\times$ the interval $[0,1]$, i.e. Cantor sets put "one next to the other" as to "cover" the quare $[0,1]\times [0,1]$ is this set self-similar, i.e. the attractor of some homotheties?
Maybe $6$ maps ... map the square $[0,1]\times[0,1]$ onto each of these:
$$
\left[0,\frac{1}{3}\right] \times \left[0,\frac{1}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[0,\frac{1}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{2}{3},1\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{2}{3},1\right]
$$
Maybe $6$ maps ... map the square $[0,1]\times[0,1]$ onto each of these:
$$
\left[0,\frac{1}{3}\right] \times \left[0,\frac{1}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[0,\frac{1}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{1}{3},\frac{2}{3}\right]
\\
\left[0,\frac{1}{3}\right] \times \left[\frac{2}{3},1\right]
\\
\left[\frac{2}{3},1\right] \times \left[\frac{2}{3},1\right]
$$