Let $\xi,\eta \in (0,\frac{1}{2})$. Let $C_\xi$ (and analogously for $C_\eta$) be the perfect symmetric set built by iterating the transformation $$[0,1] \to [0,\xi]\cup [1-\xi, 1].$$
Will the sets $C_\xi$ and $C_\eta$ be disjoint, assuming $\xi^n\neq\eta^m$ for all $n,m\ge1$?
My approach has reduced the problem to this other question.
The answer is: no, they don't have to be disjoint (even after excluding trivial common elements $0$ and $1$).
Lets fix $\xi$ and let $C_{\xi,n}$ be the $n$-th step of the iterative process of creating $C_\xi$. If $x\in C_{\xi,n}$ then it is easy to see that $\xi x\in C_{\xi,n+1}$. On the other hand if $x\in C_{\xi, n}$ then $1-x\in C_{\xi, n}$ as well, because each set in the construction is symmetric. Lets solve the equality $x=\xi(1-x)$, which gives us $x=\frac{\xi}{1+\xi}$. Denote this number by $A_\xi:=\frac{\xi}{1+\xi}$.
By induction on $n$ (and previous properties) it is easy to see that $A_\xi\in C_{\xi, n}$ for any $n$, i.e. $A_\xi\in C_{\xi}$. An interesting case is when $\xi=\frac{1}{3}$ (the standard Cantor set) for which we have that $A_\xi=\frac{1}{4}$ belongs to $C_\xi$.
Intuitively we want a number that in each step we take its symmetry $1-x$ and then scale it down by multiplying by $\xi$. If we endup in the same number each time, then we can be sure that it belongs to the Cantor-like set.
This shows that $A_\xi\in C_\xi$ and for $\eta=A_\xi$ we have $A_\xi\in C_\eta$, because it is an endpoint of $C_\eta$. And so $C_\xi$ and $C_\eta$ have a common nontrivial point. Thus the only remaining question is whether $\xi^n=\eta^m$? Such equality would imply that $(1+\xi)^m=\xi^{m-n}$. By applying $\log_{\xi}$ we can conclude that this equation doesn't have solution when $\log_{\xi}(1+\xi)$ is irrational and it does otherwise. And since $\log_\xi(1+\xi)$, as a function of $\xi\in(0,\frac{1}{2})$, is continuous and not constant then there are infinitely (even uncountably) many such $\xi$.
For a concrete example consider $\xi=\frac{1}{k}$ ($k>2$ natural) for which $\eta=A_\xi=\frac{1}{k+1}$. For those it is easy to check manually that $\xi^n\neq \eta^m$, basically because $gcd(k,k+1)=1$.
All in all, there are uncountably many $\xi$ and $\eta$ such that $\xi^n\neq \eta^m$ and $C_\xi$ has a common (non-trivial, i.e. $0$ or $1$) element with $C_\eta$. Explicitly for $\eta=\frac{\xi}{1+\xi}$.
It is an interesting question whether there even are $\xi$ and $\eta$ such that $C_\xi\cap C_\eta=\{0,1\}$. I don't know that.