It also doesn't hold on symmetric spaces in general. For example, on $T^2$ with its standard flat metric (by identifying sides of a square), if $p$ denotes the "center" point, then $D_p$ is the unique corner point while $C_p$ is the entire "boundary" square. I don't know what happens for irreducible symmetric spaces.

@PeterKropholler: Well, I suspect I am missing the key point because this answer has 7 upvotes. Your question at the top is about the lowest dimension supporting a flat manifold with perfect fundamental group. This answer addresses the lowest dimension supporting a flat manifold with perfect holonomy group. These two groups are related (perfect fundamental group obviously implies perfect holonomy group), so this answer (and Igor's comment above) provides a lower bound of $15$ for your question. But as far as I can tell, it doesn't imply the answer to your question is precisely $15$.

@PeterKropholler: I agree with everything you just wrote. The way I read Igor's answer is that we start with a flat manifold $M$ with holonomy $\phi$ with $\phi$ chosen to be perfect. Via your answer, or the Holt-Pleskin reference, we obtain a new flat manifold $N$ with holonomy $\phi$, where $\pi_1(N)$ is perfect; equivalently, $H_1(N) = 0$. One way to rephrase my question is: is it true that $\dim N \leq \dim M$? If not, I still don't see why isolating the minimal dimension in which $\phi$ can be perfect answers the question (though it does provide a lower bound).

@Peter: I am still confused, because the holonomy group doesn't determine the fundamental group, nor does it determine $H_1$. As Igor mentions in his answer in the post you linked to, if $\pi$ is the fundamental group of flat manifold $M^n$ with holonomy $\phi$, then $\pi\oplus \mathbb{Z}$ is also the fundamental group of a flat manifold $N^{n+1}$ with holonomy $\phi$, and clearly $H_1(N)\cong H_1(M)\oplus \mathbb{Z}$ in this case. Of course, minimality of the dimension in which you get holonomy $\phi$ rules out this particular construction, but why couldn't there be other constructions?

Can you expand on why minimality of dimension implies the manifold has trivial first homology?

@IanGershonTeixeira: I find your guess quite reasonable and would generally expect it to be true. But I have no idea how to prove it at this level of generality. I'm not even sure I could prove the corresponding result for $\mathbb{H}P^n = Sp(n+1)/Sp(n)\times Sp(1)$. Also, I wouldn't be surprised if there were a few examples where your method does not give the most symmetric metric, say, because that example is accidentally diffeomorphic to something else which is know to have a large symmetry group.

It is not isotropy irreducible. The group $SU(n)$ is contained in a copy of $U(n)$ lying in $SU(n+1)$. The isotropy action splits into two irreducible summands: a trivial rep (corresponding to the vectors in tangent to $U(n)$), and the usual $n$-dim complex rep of $SU(n)$.