Z-transform of x[a -n]...where a is int
There are several ways to derive the result:


Just use the definition of the $\mathcal{Z}$-transform:
$$\begin{align}\mathcal{x[a-n]}&amp;=\sum_{n=-\infty}^{\infty}x[a-n]z^{-n}{\Big|}_{k=a-n}\\
&amp;=\sum_{k=-\infty}^{\infty}x[k]z^{k-a}\\
&amp;=z^{-a}\sum_{k=-\infty}^{\infty}x[k]z^{k}=z^{-a}X\left(\frac{1}{z}\right)\end{align}$$
Use the well-known properties of the $\mathcal{Z}$-transform (version 1):
$$x[n+a]\longleftrightarrow z^aX(z)\\ [n\rightarrow -n\Rightarrow z\rightarrow 1/z]\\
x[-n+a]\longleftrightarrow z^{-a}X\left(\frac{1}{z}\right)$$
Use the well-known properties of the $\mathcal{Z}$-transform (version 2):
$$x[-n]\longleftrightarrow X\left(\frac{1}{z}\right)\\ [n\rightarrow n-a \Rightarrow \times\; z^{-a}] \\
x[-(n-a)]\longleftrightarrow z^{-a}X\left(\frac{1}{z}\right)$$