I managed to come up with an example where a connection exists. It seems to depend heavily on my choice of loss function and the use of composite hypotheses though.
I start with a general example, which is then followed by a simple special case involving the normal distribution.
General example
For an unknown parameter $\theta $, let $\Theta$ be the parameter space and consider the hypothesis $\theta\in\Theta_0$ versus the alternative $\theta\in\Theta_1=\Theta\backslash\Theta_0$.
Let $\varphi$ be a test function, using the notation in Xi'an's The Bayesian Choice (which is sort of backwards to what I at least am used to), so that we reject $\Theta_0$ if $\varphi=0$ and accept $\Theta_0$ if $\varphi=1$. Consider the loss function
$$
L(\theta,\varphi) = \begin{cases} 0, & \mbox{if } \varphi=\mathbb{I}_{\Theta_0}(\theta) \\
a_0, & \mbox{if } \theta\in\Theta_0 \mbox{ and }\varphi=0\\
a_1, & \mbox{if } \theta\in\Theta_1 \mbox{ and }\varphi=1. \end{cases}
$$
The Bayes test is then $$\varphi^\pi(x)=1\quad \rm if\quad P(\theta\in\Theta_0|x)\geq a_1(a_0+a_1)^{-1}.$$
Take $a_0=\alpha\leq 0.5$ and $a_1=1-\alpha$. The null hypothesis $\Theta_0$ is accepted if $\rm P(\theta\in\Theta_0|x)\geq 1-\alpha$.
Now, a credible region $\Theta_c$ is a region such that $\rm P(\Theta_c|x)\geq 1-\alpha$. Thus, by definition, if $\Theta_0$ is such that $\rm P(\theta\in\Theta_0|x)\geq 1-\alpha$, $\Theta_c$ can be a credible region only if $\rm P(\Theta_0\cap\Theta_c|x)>0$.
We accept the null hypothesis if an only if each $1-\alpha$-credible region contains a non-null subset of $\Theta_0$.
A simpler special case
To better illustrate what kind of test we have in the above example, consider the following special case.
Let $x\sim \rm N(\theta,1)$ with $\theta\sim \rm N(0,1)$. Set $\Theta=\mathbb{R}$, $\Theta_0=(-\infty,0]$ and $\Theta_1=(0,\infty)$, so that we wish to test whether $\theta\leq 0$.
Standard calculations give $$\rm P(\theta\leq 0|x)=\Phi(-x/\sqrt{2}),$$ where $\Phi(\cdot)$ is the standard normal cdf.
Let $z_{1-\alpha}$ be such that $\Phi(z_{1-\alpha})=1-\alpha$. $\Theta_0$ is accepted when $-x/\sqrt{2}>z_{1-\alpha}$.
This is equivalent to accepting when $x\leq \sqrt{2}z_{\alpha}.$ For $\alpha=0.05$, $\Theta_0$ is therefore rejected when $x>-2.33$.
If instead we use the prior $\theta\sim \rm N(\nu,1)$, $\Theta_0$ is rejected when $x>-2.33-\nu$.
Comments
The above loss function, where we think that falsely accepting the null hypothesis is worse than falsely rejecting it, may at first glance seem like a slightly artifical one. It can however be of considerable use in situations where "false negatives" can be costly, for instance when screening for dangerous contagious diseases or terrorists.
The condition that all credible regions must contain a part of $\Theta_0$ is actually a bit stronger than what I was hoping for: in the frequentist case the correspondence is between a single test and a single $1-\alpha$ confidence interval and not between a single test and all $1-\alpha$ intervals.
Best Answer
I said earlier that I would have a go at answering the question, so here goes...
Jaynes was being a little naughty in his paper in that a frequentist confidence interval isn't defined as an interval where we might expect the true value of the statistic to lie with high (specified) probability, so it isn't unduly surprising that contradictions arise if they are interpreted as if they were. The problem is that this is often the way confidence intervals are used in practice, as an interval highly likely to contain the true value (given what we can infer from our sample of data) is what we often want.
The key issue for me is that when a question is posed, it is best to have a direct answer to that question. Whether Bayesian credible intervals are worse than frequentist confidence intervals depends on what question was actually asked. If the question asked was:
(a) "Give me an interval where the true value of the statistic lies with probability p", then it appears a frequentist cannot actually answer that question directly (and this introduces the kind of problems that Jaynes discusses in his paper), but a Bayesian can, which is why a Bayesian credible interval is superior to the frequentist confidence interval in the examples given by Jaynes. But this is only becuase it is the "wrong question" for the frequentist.
(b) "Give me an interval where, were the experiment repeated a large number of times, the true value of the statistic would lie within p*100% of such intervals" then the frequentist answer is just what you want. The Bayesian may also be able to give a direct answer to this question (although it may not simply be the obvious credible interval). Whuber's comment on the question suggests this is the case.
So essentially, it is a matter of correctly specifying the question and properly intepreting the answer. If you want to ask question (a) then use a Bayesian credible interval, if you want to ask question (b) then use a frequentist confidence interval.