Is there distinguisher? Are these distributions computationally indistinguishable ?

$f:\{0,1\}^n \to \{0,1\}^n $

$\{X_n\}_{n\in N}$ : uniform distribution for function which $f(0^n)=0^n$  and for other function probability is zero.

$\{Y_n\}_{n\in N}$ : uniform distribution for function which $f(0^n) \neq 0^n$  and for other function probability is zero.
It looks like you're asking the following:

Consider the following families of distributions over functions from $\{0,1\}^{n} \to \{0,1\}^{n}$:

$\{X_n\}, \text{ where } X_n \text{ is the uniform distribution over the set of functions } f : \{0,1\}^{n} \to \{0,1\}^{n} \text{ such that } f(0^{n}) = 0^{n} $

$\{Y_n\}, \text{ where } X_n \text{ is the uniform distribution over the set of functions } f : \{0,1\}^{n} \to \{0,1\}^{n} \text{ such that } f(0^{n}) \neq 0^{n} $

Are these families of distributions computationally indistinguishable?

I claim they are computationally distinguishable, assuming that each function can be evaluated on input $0^{n}$ efficiently. Consider the following distinguisher, which on input $f$ computes $f(0^{n})$ and outputs $X$ if $f(0^{n}) = 0^{n}$ and $Y$ otherwise. By the definition of $\{X_n\}$, if $f$ is drawn from this family, this distinguisher will output $X$. By the definition of $\{Y_n\}$, if $f$ is drawn from this family, this distinguisher will output $Y$.