normal distribution:
Take a normal distribution with known variance. We can take this variance to be 1 without losing generality (by simply dividing each observation by the square root of the variance). This has sampling distribution:
$$p(X_{1}...X_{N}|\mu)=\left(2\pi\right)^{-\frac{N}{2}}\exp\left(-\frac{1}{2}\sum_{i=1}^{N}(X_{i}-\mu)^{2}\right)=A\exp\left(-\frac{N}{2}(\overline{X}-\mu)^{2}\right)$$
Where $A$ is a constant which depends only on the data. This shows that the sample mean is a sufficient statistic for the population mean. If we use a uniform prior, then the posterior distribution for $\mu$ will be:
$$(\mu|X_{1}...X_{N})\sim Normal\left(\overline{X},\frac{1}{N}\right)\implies \left(\sqrt{N}(\mu-\overline{X})|X_{1}...X_{N}\right)\sim Normal(0,1)$$
So a $1-\alpha$ credible interval will be of the form:
$$\left(\overline{X}+\frac{1}{\sqrt{N}}L_{\alpha},\overline{X}+\frac{1}{\sqrt{N}}U_{\alpha}\right)$$
Where $L_{\alpha}$ and $U_{\alpha}$ are chosen such that a standard normal random variable $Z$ satisfies:
$$Pr\left(L_{\alpha}<Z<U_{\alpha}\right)=1-\alpha$$
Now we can start from this "pivotal quantity" for constructing a confidence interval. The sampling distribution of $\sqrt{N}(\mu-\overline{X})$ for fixed $\mu$ is a standard normal distribution, so we can substitute this into the above probability:
$$Pr\left(L_{\alpha}<\sqrt{N}(\mu-\overline{X})<U_{\alpha}\right)=1-\alpha$$
Then re-arrange to solve for $\mu$, and the confidence interval will be the same as the credible interval.
Scale parameters:
For scale parameters, the pdfs have the form $p(X_{i}|s)=\frac{1}{s}f\left(\frac{X_{i}}{s}\right)$. We can take the $(X_{i}|s)\sim Uniform(0,s)$, which corresponds to $f(t)=1$. The joint sampling distribution is:
$$p(X_{1}...X_{N}|s)=s^{-N}\;\;\;\;\;\;\;0<X_{1}...X_{N}<s$$
From which we find the sufficient statistic to be equal to $X_{max}$ (the maximum of the observations). We now find its sampling distribution:
$$Pr(X_{max}<y|s)=Pr(X_{1}<y,X_{2}<y...X_{N}<y|s)=\left(\frac{y}{s}\right)^{N}$$
Now we can make this independent of the parameter by taking $y=qs$. This means our "pivotal quantity" is given by $Q=s^{-1}X_{max}$ with $Pr(Q<q)=q^{N}$ which is the $beta(N,1)$ distribution. So, we can choose $L_{\alpha},U_{\alpha}$ using the beta quantiles such that:
$$Pr(L_{\alpha}<Q<U_{\alpha})=1-\alpha=U_{\alpha}^{N}-L_{\alpha}^{N}$$
And we substitute the pivotal quantity:
$$Pr(L_{\alpha}<s^{-1}X_{max}<U_{\alpha})=1-\alpha=Pr(X_{max}L_{\alpha}^{-1}>s>X_{max}U_{\alpha}^{-1})$$
And there is our confidence interval. For the Bayesian solution with jeffreys prior we have:
$$p(s|X_{1}...X_{N})=\frac{s^{-N-1}}{\int_{X_{max}}^{\infty}r^{-N-1}dr}=N (X_{max})^{N}s^{-N-1}$$
$$\implies Pr(s>t|X_{1}...X_{N})=N (X_{max})^{N}\int_{t}^{\infty}s^{-N-1}ds=\left(\frac{X_{max}}{t}\right)^{N}$$
We now plug in the confidence interval, and calculate its credibility
$$Pr(X_{max}L_{\alpha}^{-1}>s>X_{max}U_{\alpha}^{-1}|X_{1}...X_{N})=\left(\frac{X_{max}}{X_{max}U_{\alpha}^{-1}}\right)^{N}-\left(\frac{X_{max}}{X_{max}L_{\alpha}^{-1}}\right)^{N}$$
$$=U_{\alpha}^{N}-L_{\alpha}^{N}=Pr(L_{\alpha}<Q<U_{\alpha})$$
And presto, we have $1-\alpha$ credibility and coverage.
