Did I solve $z^{n-1}=\overline{z}$ correctly? 
Please could someone tell me if my solution of $z^{n-1}=\overline{z}$
  is correct?

My solution:
We have $z^{n-1} =\overline{z}$ if and only if $z^{n-1}z =\overline{z}z$, so
$z^n = |z|^2$
In polar coordinates, 
$$ r^n e^{i n \varphi}  = r^2$$
or, equivalently, 
$$ r^{n-2} e^{i n \varphi}  = 1$$
This equation implies that $r=1$. Hence we are taking $n$-th roots of $1$:
$$e^{i n \varphi}  = 1$$
So the resulting set is a discrete and finite subset of the unit circle (consisting of the $n$-th roots of $1$). 

I assume $n$ stands for a positive integer.
If $n=1$, the equation is $1=\bar{z}$.
If $n=2$, the equation is $z=\bar{z}$, so any real number is a solution.
Suppose $n>2$. First of all, $z=0$ is a solution. Assume now $z\ne0$ and write it as $z=re^{i\varphi}$. The equation becomes
$$
r^{n-1}e^{i(n-1)\varphi}=re^{-i\varphi}
$$
Since $r>0$, we can write it as
$$
r^{n-2}e^{in\varphi}=1
$$
Since $n>2$, we get $r=1$ and $e^{in\varphi}=1$, so the $n$-th roots of unity are the solutions.

If $n$ is allowed to be any integer, the case $n=0$ becomes $z\bar{z}=1$, so any complex number of modulus $1$ is a solution.
If $n<0$, write $w=1/z$ and $m=-n$, so the equation becomes
$$
\frac{1}{w^{m+1}}=\frac{1}{\bar{w}}
$$
or
$$
w^{m+1}=\bar{w}
$$
and we're in the same situation as before with $n>2$ (but, of course, the solution $0$ cannot be considered).
