Here is a solution with
$5$ rooks $(6, 14, 26, 38, 50)$:

I don't think less can be possible since
$4$ are required to attack the ends of the horizontals which together cannot attack all other cells. See how Yellow is required to attack $(2,3,31,34)$ in the above solution.
Update
Here is code to search for solutions
- it is not generalised to any size, I just hard-coded the "rows" (horizontals, diagonals and anti-diagonals)
- it is zero based so all the numbers are one less that those shown in the question
N_CELLS = 57
ROWS = [{4,5,6,7,8,9,10,11},{19,20,21,22,23,24,25,26},{34,35,36,37,38,39,40,41},{49,50,51,52,53,54,55,56}, # Horizontals
{50,45},{42,35,52,46,30},{27,37,43,15,20,54,47,31},{32,0,48,5,39,12,44,16,22,56,28},{33,1,41,7,13,17,24,29},{9,18,26,2,14},{11,3}, # Diagonals
{0,4},{1,19,12,6,15},{34,8,2,13,16,21,27,30},{3,36,49,10,45,14,17,23,28,42,31},{32,38,43,46,18,51,25,29},{40,33,44,53,47},{48,55} # Anti-diagonals
]
ALL_CELLS = set(range(N_CELLS))
def makeAttackLookup():
res = []
for pos in range(N_CELLS):
a = set()
for r in ROWS:
if pos in r:
a |= r
res.append(set(a))
return res
ATTACK_LOOKUP = makeAttackLookup()
def iterSolutions(stop, curPos=[], curAttacks=set()):
if len(curAttacks) == N_CELLS:
yield curPos
elif len(curPos) < stop:
for pos in ALL_CELLS - curAttacks:
if not curPos or pos > curPos[-1]:
for solution in iterSolutions(stop, curPos + [pos], curAttacks | ATTACK_LOOKUP[pos]):
yield solution
Counting solutions (not collapsing into equivalence classes):
>>> c = 0
>>> for solution in iterSolutions(4): c+=1
...
>>> c
0
>>> for solution in iterSolutions(5): c+=1
...
>>> c
162
We can count how many there are up to symmetry by noting that
All solutions of $5$ must use exactly one of the top horizontal that is not an end: $\{6,7,8,9,10,11\}$
A solution using $6$ has a reflection in the vertical using $11$
Likewise for $7,10$ and $8,9$
There are $29$ solutions using $6$, $14$ using $7$ and $38$ using $8$
Total number of solutions up to symmetry is $29+14+38=81$ (half the number counted as expected by the fact there is only one reflection in the board).