Effect of sign change on ROC & z transform?
You need to understand that a function such as $F(z)=\frac{z}{z+1}$ doesn't have a region of convergence (ROC). Only an infinite series has a ROC. E.g., the $\mathcal{Z}$-transform of a sequence $x[n]$ can be expressed by the series

$$\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$

and this expression only makes sense if the series converges for certain values of $z$. The ROC is the region of the $z$-plane for which the series $(1)$ converges.

If you turn the problem around and if you're given a function of $z$ like $F(z)$ defined above, you can ask yourself if $F(z)$ is a valid $\mathcal{Z}$-transform of some sequence $f[n]$. It turns out that $F(z)$ is the $\mathcal{Z}$-transform of two different sequences, but with different ROCs. So a $\mathcal{Z}$-transform is only completely defined by the expression for the function $and$ the corresponding ROC.

Coming back to the function $F(z)$, it's straightforward to show that it is the $\mathcal{Z}$-transform of the sequence $f_1[n]=(-1)^nu[n]$ with the ROC $|z|&gt;1$, and of the sequence $f_2[n]=-(-1)^nu[-n-1]$ with the ROC $|z|&lt;1$.

Concerning your last question, as mentioned above, the expression for the $\mathcal{Z}$-transform including the ROC is unique, so the two sequences $u[n]$ and $u[-n]$ cannot have the same $\mathcal{Z}$-transform and ROC. It's actually pretty easy to come up with the $\mathcal{Z}$-transform of $u[-n]$:

$$\begin{align}\mathcal{Z}\big\{u[-n]\big\}&amp;=\sum_{n=-\infty}^0z^{-n}\\&amp;=\sum_{n=0}^{\infty}z^n\\&amp;=\frac{1}{1-z},\qquad |z|&lt;1\end{align}$$

which is of course different from the $\mathcal{Z}$-transform of $u[n]$. Note that the algebraic expression as well as the ROC are different.