Intro
When looking to pictures of the generalize Mandelbrot one is tempted to make at least two observations. First, that the set seems to be always symmetric around the real axis. Second, the sets are symmetric under angle rotation. We will quickly show that both are indeed true.
Sets for k=2, 5 and 8, drawn with a distance estimation algorithm
Some pre math
We are going to concentrate on the Mandelbrot sets for monic one dimensional polynomials of the form . Note that the critical point is the same
for all of them (zero), and that infinity is an superattractive fixed point. The Mandelbrot set is defined as for the standard case k=2. The set of point created by the iteration of with = 0 is called the orbit of . Each point in the orbit is thus and when applied to = 0 it expands the next polynomials in c: . So, and so on. Then, a point c belongs to if stays bounded as 
The Mandelbrot set for k=4 

Vertical Symmetry
Vertical symmetry translates to saying that if a given point c belongs to the set, then it's conjugate does also: . The result comes from the fact that the iteration of develops a polynomial in c with only real coefficients. But, let's show it anyway by induction.
Let's assume that for one of the iterations we have
meaning that if c belongs to then will also do, since the modulus of a complex number and its conjugate is the same. So, this is symmetrical. Let's examine now what happens to the next iteration both for c
and for
From the property that for any complex number w and integer a we have , we conclude
and thus the next iterate is symmetrical too. In other words
We just need to check that is symmetric to arrive to the conclusion that all the iterates are symmetrical.
Therefore the complete set is symmetrical around the real axis:
Rotational Symmetry
For the rotational symmetry we proceed in a similar way. Let's call a nsymmetry to a rotational symmetry of radians, for any integers m and n. Now, we first assume that one iterate is (k1)symmetric as the pictures suggest. That translates to
and now we try to demonstrate that next iterate will also be symmetrical. For that we analyze the iteration:
By the assumption we made,
So, again,
Because we can easily check that for n=1 the assumption holds, and because we can say that
Or put it in other way,