Assume we pick uniformly random a permutation polynomial, $T$, of degree one. we define all polynomials over $\mathbb{Z}_P$. We have fixed inputs $x_i$ (e.g. $x_i \in [1,100]$)
- My Question: Is $y_i=T(x_i$) a uniformly random element of $\mathbb{Z}_p$.
Let $p$ be a large prime number.
If your random permutation polynomials are of degree one, they have the form $T(x) = ax+b$ for some $a, b \in \mathbb{Z}_{p}$, $a \neq 0$. I'm not sure how many fixed inputs there are. For a single input $x_{1}$, $T(x_{1}) = ax_{1}+b = y_{1}$ whenever $b = y_{1}-ax_{1}$; there is precisely one $b$ for any choice of $a$, and $p$ choices for $b$, so the probability is $1/p$ (so uniform).
Now, it is not really clear from your question how many fixed inputs you are evaluating, but if there are more than two than the collection of outputs cannot be uniformly distributed because your function $T$ is always linear. That means two outputs determines all of the outputs.