Can permutating the digits of an irrational/transcendental number give any other such number? Let $x_n$ be the infinite sequence of decimal digits of a fixed irrational/trascendental number. Can I obtain any other irrational/trascendental number's sequence of decimal digits through a permutation of $x_n$?
 A: In base 2, yes.  Any irrational number (algebraic or transcendental) will have an infinite number of 1s and 0s.
Reductio ad absurdum, assume either 1's or 0'1 are finite.  After the last of the finite digit types is written down you'd be left with 0111111111111... (which can be re-written as 1) or 1000000000000... (where you can discard the infinite zero trail).  Ergo your number would be rational, contradiction.  So the assumption led to contradiction, q.e.d.
So the original number provides an infinite set of 0s and an infinite set of 1s. From these two infinite sets, you can at leisure reconstruct any transcendental, algebraic, or even rational number. 

In any other base, however, no.  As there are more than two options for digits, the original number may have only a finite number of a certain digit (and an infinite of the other 2 or more).  So any number that needs more digits of a certain type than your number has in store will fail.  
For example, in base 3, one can construct an uncountable set of numbers using only digits 0 and 2 (take all the irrational numbers written in base 2, replace 1 with 2).  An irrational number written with 0s and 2s only cannot be permutated into an irrational number that has 1s in its expansion.
