Yes, $\bmod 10\,$ the cube map $\,c(n) = n^3$ is a bijection by $\, a \equiv a^9 = (a^3)^3 = c(c(a))\,$ therefore $\,c\,$ is self-inverse. Indeed $\,a^9\equiv a\pmod{\!10}\,$ follows by a slight generalization of little Fermat / Euler.
You can observe this in your table: $\,2\mapsto 8\mapsto 2,\,$ $\, 3\mapsto 7\mapsto 3\,$ and others are fixed points so $\,c(c(a)) = a,\,$ i.e. $\,c^2 = 1,\,$ so $\,c^{-1} = c,\,$ i.e. $\,c\,$ is the permutation $\,(2\ 8)\,(3\ 7)\,$ in cycle notation.
More generally if $\,n\,$ is a product of distinct primes $\equiv 2\pmod{\!3}\,$ then $\,\phi = \phi(n)\equiv 1\pmod{\!3}.\,$ Hence $\bmod 3\!:\ \phi\equiv 1\,\Rightarrow\,2\phi+1\equiv 0\,$ so $\,\color{#c00}{2\phi+1 = 3k}\,$ so the above linked theorem yields
$$\bmod n\!:\,\ 0\equiv a(a^{2\phi}-1)\,\Rightarrow\,a\equiv (a^{\large\color{#c00}k})^{\large 3}\equiv a^{\large \color{#c00}{2\phi+1}} \qquad\qquad\!\!\!$$
e.g. $\, \begin{align} n &= 2\cdot 5\cdot 11 = 110\\ \phi &= 1\cdot 4\cdot 10 = 40\end{align}\ $ so $\,2\phi+1 = 27\cdot 3\ $ so $\ a\equiv (a^{\large 27})^{\large 3}\pmod{\!110}$
When $\,n\,$ is a prime $\equiv 2\pmod{\!3}\,$ the above specializes to a simple case of cubic reciprocity.
For the 2nd question, by Fermat, $\!\bmod 7\!:\ n\not\equiv 0\,\Rightarrow 0\equiv n^6-1 = (n^3-1)(n^3+1)$ so $7$ divides one of the factors by Euclid's Lemma, so $\,n^3\equiv \pm1.\,$ More generally see Euler's Criterion.
