Would two different pivotal quantities of the same parameter give the same confidence interval? I couldn't think of an example but it would be great if someone could give one.
 A: If $U(X,\theta)$ is a pivotal quantity, then so is  $g(U(X,\theta))$, where $g(\cdot)$ does not involve the parameter. Therefore, the there are literally infinite numbers of pivots, but not all of them will be of the same length for a given confidence level.
Revised per OP's Comments
As a specific example, lets take the familiar confidence interval for the mean given a sample of size $N$ from a normal population with known standard deviation $\sigma$:
We know from basic statistics that $\frac{\sqrt{N}(\bar X-\mu)}{\sigma}\sim \mathcal{N}(0,1)$, which is free of $\mu$ and therefore is a pivotal quantity for our particular inference problem. 
Now, lets take $g(z)=z^2$ as our parameter-free function. If we apply it to the above pivot, we get:
$g\left(\frac{\sqrt{N}(\bar X-\mu)}{\sigma}\right)\sim \chi^2_1$, which is also a pivotal quantity.
If we form a 95%CI from the above two pivotal quantities, we get:
$\bar X \pm \frac{1.96 \sigma}{\sqrt{N}}$ for the familiar "textbook" CI based on a normally distributed pivotal quantity. 
For the one based on the Chi-squared pivotal quantity, we have to do a little more work:
$P\left(\chi^2_{(.025,1)}\leq \left(\frac{\sqrt{N}(\bar X-\mu)}{\sigma}\right)^2 \leq \chi^2_{(.975,1)}\right) = 0.95 \rightarrow$
$P\left(\sigma\sqrt{\frac{\sigma \chi^2_{(.025,1)}}{N}} \leq |\mu -\bar X| \leq \sigma\sqrt{\frac{\sigma \chi^2_{(.975,1)}}{N}}\right)=0.95 \rightarrow$
$P\left(\left[\sigma\sqrt{\frac{\chi^2_{(.025,1)}}{N}} \leq \mu-\bar X \leq \sigma\sqrt{\frac{\chi^2_{(.975,1)}}{N}}\right]\cup \left[-\sigma \sqrt{\frac{\chi^2_{(.975,1)}}{N}} \leq \mu-\bar X \leq \sigma \sqrt{\frac{\chi^2_{(.025,1)}}{N}}\right]\right)=0.95 \rightarrow$
$P\left(\left[\bar X+\sigma \sqrt{\frac{\chi^2_{(.025,1)}}{N}} \leq \mu \leq \bar X+\sigma \sqrt{\frac{\chi^2_{(.975,1)}}{N}}\right]\cup \left[\bar X-\sigma \sqrt{\frac{\chi^2_{(.975,1)}}{N}} \leq \mu \leq \bar X-\sigma \sqrt{\frac{\chi^2_{(.025,1)}}{N}}\right]\right)=0.95$
$\square$
Hence, our "CI" in this case is really a confidence set, since it is composed of two disjoint intervals: $\left[\bar X+\sigma \sqrt{\frac{\sigma \chi^2_{(.025,1)}}{N}} \leq \mu \leq \bar X+\sigma \sqrt{\frac{\chi^2_{(.975,1)}}{N}}\right]\cup \left[\bar X-\sigma \sqrt{\frac{\chi^2_{(.975,1)}}{N}} \leq \mu \leq \bar X-\sigma \sqrt{\frac{ \chi^2_{(.025,1)}}{N}}\right]$
This interval is not only qualitatively different from the "typical" CI, but it is also wider, and hence not the optimal interval (we usually want the shortest interval of a given confidence).
A: I don't have the enough reputation to comment on the chi-square derived interval/set below, but it appears to me that the true parameter is not even in the set described: $C(X) = (\bar{X}-c2,\bar{X}-c1) \cup (\bar{X}+c1,\bar{X}+c2)$
It should still be an interval given by the union $C(X) = (\bar{X}-c1,\bar{X}+c1) \cup (\bar{X}-c2,\bar{X}+c2)$ where $c1=\chi^2_{0.025,1}, c2=\chi^2_{0.975,1}$.
A: Assume that you observe iid observations $X_1,...,X_n$ from some distribution, for the sake of simplicity let's say they are normal $N(\theta,1)$ for some unknown parameter $\theta$ that you wish to estimate.
By the central limit theorem, a first pivotal quantity
is $$\frac{1}{\sqrt n} \sum_{i=1}^n (X_i - \theta) \sim N(0,1).$$
But the central limit theorem also yields plenty of other pivotal quantities
by applying transformation to each $X_i$, for example
$$\frac{1}{\sqrt{2n}} \sum_{i=1}^n ((X_i-\theta)^2-1) \Rightarrow N(0,1)$$
because $(X_i-\theta)^2$ has mean 0 and variance 2.
How to pick the right pivotal quantity then?
All pivotal quantities are not created equal, and since several (many!) pivotal quantities can be obtained by the central limit theorem, it would be great to have some principled way to do that. In frequentist statistics, where you assume that the dataset comes from a distribution with an unknown parameter $\theta$ that you wish to estimate, two major principles come to mind:

*

*sufficient statistic, see https://en.wikipedia.org/wiki/Sufficient_statistic and the related optimality results, for instance https://en.wikipedia.org/wiki/Minimum-variance_unbiased_estimator.


*Maximum-likelihood estimation. If you you know the unknown distribution comes from a parametric family of distributions indexed by an unknown parameter $\theta$, you should start by looking at the Maxmimum-Likelihood estimate of the problem, which satisfies general asymptotic normality properties, and is often optimal in the sense that it achieves the Cramer-Rao lower boud. See https://en.wikipedia.org/wiki/Maximum_likelihood_estimation
