Algebra on random variables
You can calculate the cdf. Let $Z=-2X+1$. Then
$P(Z<z)=P(-2X+1<z)=P(1-z<2X)=P\left(\frac{1-z}{2}<X \right)$
The cdf of X is
$$F_X(x)=\begin{cases}0, \text{if} \ x<0 \\ x, \text{if} \ 0 \leq x<1 \\ 1, \text{if} \ x\geq 1 \end{cases}$$
Then use $P\left(\frac{1-z}{2}<X \right)=1-P\left(X<\frac{1-z}{2} \right)$. And you have to adjust the limits: $x=0 \Rightarrow z=1, x=1\Rightarrow z=-1$
Therefore the cdf of $Z$ is
$$F_Z(z)=\begin{cases}0, \text{if} \ z<-1 \\ 1-\frac{1-z}{2}, \text{if} \ -1 \leq z<1 \\ 1, \text{if} \ z\geq 1 \end{cases}$$
Is Z uniformly distributed on $(-1,1)$ ?