Do not ask me what this is because I do not know but... would not fall solutions on intersection of surfaces?

R = .5;
Plot3D[{Im[(a + I b)^(1/(a + I b))], Re[(a + I b)^(1/(a + I b))]}, {a, -R, R}, {b, -R, R}, 
Mesh -> None, PlotPoints -> 100, ExclusionsStyle -> Opacity[.5]]

Another way - looks a bit like Mandelbrot set:

R = .01;
ContourPlot[Im[(a + I b)^(1/(a + I b))] == Re[(a + I b)^(1/(a + I b))], {a, -R, R}, {b, -R, R}, 
PlotPoints -> 50, ContourStyle -> Directive[Thickness[0], Opacity[.5]]]

The roots of x^(1/x)==k are functions of Log and ProductLog.

x /. Solve[(1/5)^x == x, x, Reals]

(* Out[196]= {ProductLog[Log[5]]/Log[5]} *)

To get a better sense of what goes on with the re/im of x^(1/x) when x is between -1 and 0, I'd recommend starting with ComplexExpand. Also simplify using that range.

ss = Simplify[ComplexExpand[Re[x^(1/x)] - Im[x^(1/x)]],  Assumptions -> -1 < x < 0]

(* Out[199]= (-x)^(1/x) (Cos[\[Pi]/x] - Sin[\[Pi]/x]) *)

Now solve for this vanishing and simplify the result.

FullSimplify[x /. Solve[ss == 0 && -1 <= x < 0, x]]

(* Out[202]= {ConditionalExpression[4/(1 + 8 C[1]), 
  C[1] \[Element] Integers && C[1] <= -1], 
 ConditionalExpression[4/(-3 + 8 C[1]), 
  C[1] \[Element] Integers && C[1] &lt;= -1]} *)

So those denominators are comprised of two arithmetic sequences which, negating to put them in the more customary form, are 8*n-1 and 8*n+3. That is, {7,15,23,31,39,47,55,...} and {11,19,27,35,43,51,...}. Note that 3 is ruled out by the original restriction that -1<=x&lt;0

Anyway, here are a couple of immediate observations. One is that the denominators that "work" are not all prime. The other is that the primes that "fail" are those of the form 8n+1 (e.g. 17,41,...) and 8n+5 (so 13,29,...).

In like manor, for the function g(x)=(-1)^x *x^(1/x) in (0,Infinity), Im[g[x]=Re[g[x] at rational values also!

In[322]:= g[x_] = (-1)^x *x^(1/x)

Out[322]= (-1)^x x^(1/x)

In[324]:= Plot[Im[g[x]] - Re[g[x]], {x, 0, 10}]

In[326]:= FindRoot[Im[g[x]] - Re[g[x]], {x, {2, 3, 4, 5, 6, 7}}]

Out[326]= {x -> {2.25, 3.25, 4.25, 5.25, 6.25, 7.25}}

In[334]:= FullSimplify[
 Table[{q = x + 1/4; Im[g[q]] == Re[g[q]]}, {x, 1, 100}]]

Out[334]= {{True},..., {True}}