How to prove independence in the following case? Let $f$ be a (possibly randomized) function with finite domain $D$ and finite range $R$, such that for any two $x$ and $x'$ in the domain, $f(x)$ and $f(x')$ are identically distributed; i.e.:
$$\forall x,x'\in D, \forall y\in R \quad \Pr[f(x)=y] = \Pr[f(x')=y] \enspace. \qquad (\dagger)$$
I want to prove that the output of $f$ is independent of any input distribution. That is, for any discrete input distribution $X$ with support $D$, we have:
$$\forall x\in D, \forall y\in R \quad \Pr[f(x)=y \mid X=x]=\Pr[f(x)=y] \enspace. \qquad (\ddagger)$$

Here's two examples that satisfy $(\dagger)$, and therefore satisfy $(\ddagger)$:
Example 1. $f(x)$ is a constant function.
Example 2. Domain $D$ is the set of $n$-bit strings, and $f(x)=x \oplus r$. That is, $f$ picks a random $n$-bit string (internally), and XORs it to its input.
 A: Definitions and clarification
Let $f$ be a randomized function from $D$ to $R$.  In the context of the question this implies that for each $x\in D$, $f$ assigns a probability $f(x,y)$ to each $y\in R$. (Actually, it does not matter that $f(x,y)$ is a probability: it merely has to be any element of a real vector space, such as a real number or tuple of real numbers.  Probabilities are real numbers.)
Suppose $p$ is a probability distribution over $D$ (entailing, in particular, that $\sum_{x\in D}p(x)=1$; it turns out this is all we need to know about $p$).  Then--this is how I interpret the question--$p$ acts on $f$ by mixing it to create a probability distribution $f^{(p)}$ on $R$ where, for any $y\in R$,
$$f^{(p)}(y) = \sum_{x\in D} p(x) f(x,y).$$
(When all the $f(x,*)$ are probability distributions and $p$ is itself a probability distribution, then in fact $f^{(p)}$ is itself a probability distribution: this is straightforward to show.  But that doesn't matter for this demonstration.)
Solution
Suppose $f(x,y)$ is independent of $x$ (which is the assumption of the question); that is, we may write $f(x,y) = g(y)$ for some function $g$.  Then
$$f^{(p)}(y)  = \sum_{x\in D} p(x) f(x,y) = \sum_{x\in D} p(x) g(y) =  g(y)\sum_{x\in D}p(x) = g(y)1 = g(y),$$
QED.
A: I will compact notation a bit. We have a discrete support, $D=\{x_1,...,x_n\}$. We define the events $f_i:\{f(x_i)=y\}, \; f_k:\{f(x_k)=y\},\; i,k =1,...,n$
Then, for any given $y$, your premise can be written as  $$\Pr(f_i) = \Pr(f_k)\; \forall i,k$$
and you want to show that 
$$\Pr(f_i|x_i) = \Pr(f_i)\; \forall i$$
By Bayes theorem
$$\Pr(f_i) = \sum_i^n\Pr(f_i|x_i)\Pr(x_i) $$ $$= \Pr(f_i|x_1)\Pr(x_1)+...+\Pr(f_i|x_i)\Pr(x_i)+...+\Pr(f_i|x_n)\Pr(x_n)$$
But $\Pr(f_i|x_k) = 0 \;\;\forall\, k\neq i$. If the realized value of $X$ is $x_k\;, k\neq i$, the event ${f(x_i)=y}$ is impossible, because $f(x_i)$ is impossible - only $f(x_k)$ is possible  . So we are left with
$$\Pr(f_i) =  \Pr(f_i|x_i)\Pr(x_i)$$. 
So what we wanted to prove is not correct. But it seemed fairly intuitive... There is another relation here that I believe represents this intuition. Use a prime to denote that all previous analysis is conducted for the probability that $f$ takes the value $y'$. Then we will arrive at 
$$\Pr(f'_i) =  \Pr(f'_i|x_i)\Pr(x_i)$$
Combining the two results we obtain
$$\frac {\Pr(f_i)}{\Pr(f_i|x_i)} = \frac {\Pr(f'_i)}{\Pr(f'_i|x_i)} $$
$$\Rightarrow \frac {\Pr(f_i)}{\Pr(f'_i)} = \frac {\Pr(f_i|x_i)}{\Pr(f'_i|x_i)} $$
Moreover, since $\Pr(f_i)=\Pr(f_k)$ and $\Pr(f'_i)=\Pr(f'_k)$ we obtain
$$ \frac {\Pr(f_i)}{\Pr(f'_i)} = \frac {\Pr(f_i|x_i)}{\Pr(f'_i|x_i)} =\frac {\Pr(f_k|x_k)}{\Pr(f'_k|x_k)}=\frac {\Pr(f_k)}{\Pr(f'_k)}$$
The key here is that the realization of some $x_i$ may affect the distribution of $f(x)$ -but it will do it in a way that leaves relative probabilities unchanged.
