Question
Consider some function $f(v)$ of a complex vector $v \in \mathbb{C}^N$. Why is averaging $f(v)$ over complex vectors drawn iid from the uniform distribution over the unit ball $|v|=1$ not the same as averaging $f(Ux)$ over unitaries $U$ drawn from the Haar invariant measure on unitaries of dimension $N$, for some fixed unit vector $x \in \mathbb{C}^N$, $|x|=1$?
As both of these measures are defined with a notion of rotational invariance, the result that they are different seems very counterintuitive to me. Is there an intuitive explanation for it?
Specific example
I was interested to perform the following average over unit complex vectors $$ K = \frac{\int_{v \in \mathbb{C}^N} \delta(|v|-1) d v \exp\left({v^\dagger A v}\right)}{\int_{v \in \mathbb{C}^N} \delta(|v|-1) d v} $$ where $A$ is a $d \times d$ hermitian matrix.
My intuition was that I could treat this as a special case of the more general Itzykson-Zuber integral $$ I = \int_{U(N)} d U \mathrm{e}^{\operatorname{tr}(A U B U^\dagger)} = \left(\prod_{p = 1}^{N-1} p! \right) \frac{\mathrm{det}\left(\mathrm{e}^{a_i b_j}\right)}{\Delta(a)\Delta(b)} $$ where in this special case $B$ is a matrix with a single non-zero eigenvalue equal to one. In the relation above $a_i$ and $b_i$ are the eignevalues of $A$ and $B$ respectively, and $\Delta(a) = \prod_{1 \leq j < i \leq N}(a_i - a_j)$ is the Vandermonde determinant, and the integral is taken over the Haar invariant measure on unitary matrices of dimension $N$.
Setting $b_1 = 1$, and taking the limit $b_{n>1} \to 0$. One obtains $$ I' = \int_{U(N)} d U \mathrm{e}^{x^\dagger U^\dagger A U x} = (N-1)! \sum_{n=1}^N \frac{\mathrm{e}^{a_n}}{\prod_{m \neq n}(a_n - a_m)} $$ where $x$ is a fixed complex vector of unit length.
However on numerically verifying my calculation, I was surprised to find $$ K \neq I' $$ why is this?
Specifically, for $A = \mathrm{diagonal}(1,2,3,4)$ I draw random complex numbers $z = x + i y$ where $x,y \sim \text{Normal}(0,1)$ are iid, and normalise them to provide a random complex vector $v$ satisfying $|v|=1$, and I then calculate $\exp\left({v^\dagger A v}\right)$. On repeating this process many times, I find a sample mean which converges to $$ K \approx 14.954 $$ In contrast, when I fix $x = (1,0,0,0)$, drawing unitary $U \sim \text{CUE}(N)$, and calculate $\mathrm{e}^{x^\dagger U^\dagger A U x}$ I find a sample mean which converges to $$ I' = \mathrm{e}^4 -3 \mathrm{e}^3 +3 \mathrm{e}^2 -\mathrm{e} = 13.7904... $$ as expected.