How to generalize the Thue-Morse sequence to more than two symbols? The Thue-Morse sequence is defined as a binary sequence and can be generated like  
  0, 01, 01 10, 01 10 10 01, 01 10 10 01 10 01 01 10, ... . 

So the second half of the series is always the binary complement of the first half of the series. 
But is there a way to generate an analogous ternary sequence?
Intuitively my first guess for a ternary Thue-Morse sequence was like
  0, 01, 012, 012 120, 012 120 120 201, 012 120 120 201 120 201 201 012, ...

So here the second half of the series is the "ternary complement" (rotation $0 \to 1, 1 \to 2, 2 \to 0 $ instead of $0 \to 1, 1 \to 0 $) of the first half. 
But it could also be
  0, 01, 012, 012 120 201, 012 120 201 120 201 012 201 012 120, ... 

Here the second third is the "ternary complement" of the first third and the third third is the "ternary complement" of the second third. 

Does any of my constructions for a ternary Thue-Morse series make sense?
  Is there maybe a unique way to generate an analogous ternary sequence?
  And how to construct n-ary versions of the Thue-Morse series in general?

 A: To find out a particular number n for b symbols, convert the number to base b, sum the digits, and then modulo by b.
For example, you have two symbols, 0 and 1, and you want to find out what the fifteenth place in this sequence would be. Convert 14 (14 is the fifteenth number since this sequence starts with 0) to base two (the number of symbols), it's 1110. Sum the digits, it's 3. Take 3 modulo two (again, the number of symbols). It's 1. So the fifteenth place in the binary Thue-Morse sequence is 1.
If you have four symbols, 0 and 1 and 2 and 3, and you want to find out the twelfth place in the sequence. Convert 11 to base four, it's 23. Sum the digits, it's 5. Five modulo four is 1, so the twelfth place in the quaternary Thue-Morse sequence is also 1.

That was my own discovery back in 2016.
I knew about
T(2n) = T(n)
T(2n+1) = not T(n)

I saw that
T(2n+1) = not T(n) = (T(n)+1) modulo 2.

and thought that one possible generalization for other numbers of symbols could be:
T(bn) = T(n)
T(bn+y) = (T(n)+y) modulo b

where y is lower than b.
Here is a shellscript that prints 0 to $1 iterations where $p is the amount of symbols.
#!/bin/sh
p=6
for i in `seq 0 $1`
do
    echo -n "$(echo "${p}o${i}f"|dc|sed 's/\(.\)/\1 /g')  [+lbx]sa[2z>a]dsbx${p}%n"|dc
done
echo

