If you know a normal distribution's population variance, does a sample variance tell you nothing about the sample's mean's confidence interval? If I understand correctly (which I might not), if I know a normal distribution's population variance but not its population mean, and take just one sample consisting of three measurements, then no matter what my sample variance is I should calculate 95% Confidence Intervals from the population variance, in the form $1.96*\sqrt{\frac{[PopulationVariance]}{3}}$, and that these can be used as error bars about the sample mean.
My question:  does the sample variance tell me nothing about the confidence interval?  If considering all sets of three measurements that I could take for a normal distribution, is there no overall correlation between low sample variance and nearness to the mean?  
(Based on the thinking that more possible measurements are near the middle, so measurements similarly spaced but further from the mean would be less likely.)
If the sample variance does tell me something about what the confidence interval should be, how do I incorporate this without estimating a different population variance estimate from scratch from the sample variance?
Further thoughts following Xi'an's response:  In a spreadsheet (from values -50 to 50), taking a normal distribution X of standard deviation 10 and mean 0 (representing the unknown population mean $\mu$), I can then multiply offset probabilities of X with itself; for a distance of 5 between two measurements, I can for instance get the probability that the lower measurement is at $\mu$ and that the higher measurement is at $\mu+5$ through $p(Z=0) = p(X=0) * p(X=5)$.  I then compared the probability distributions for distances of 5 and 1, with mid-points shifted to 0 and using different y-axes, and saw that the distributions indeed appear to be exactly overlaid (without different variance)!  I now no longer suspect that measurement-clustering gives information about population-mean closeness, but at time of writing do not yet intuitively understand why.
Yet another thought/note:  (If I understand correctly, the integral of the product of two probability density functions (each with integral 100% probability) is not itself 100%, requiring scaling, so a lower central probability does not imply higher tail probabilities.)
 A: 
An ancillary statistic is a measure of a sample whose distribution
  does not depend on the parameters of the model. An ancillary statistic
  is a pivotal quantity that is also a statistic. Ancillary statistics
  can be used to construct prediction intervals. Wikipedia

For a Normal sample $X_1,\ldots,X_N\sim\mathcal N(\mu,\sigma^2)$ where $\sigma$ is known, a minimal sufficient and complete statistic is $\bar X_n$, while the sample variance $s^2_n = \sum_{i=1}^n (X_i-\bar X_n)^2 / n$ is ancillary. While a general approach for constructing confidence intervals on $\mu$ is to condition on $s^2_n$, the special case of the Normal distribution means that $\bar X_n$ is independent from $s^2_n$ and hence $s^2_n$ cannot be used to improve the length of the confidence interval on $\mu$.
A: The short answer is "yes, nothing" 8-)
The reason is easiest to see if you take just two samples from a distribution with sample variance 1.0.  If we see samples of 3.9 and 4.1 we know we have just been lucky that they are that close.  We can't do any better than 4.0 as our estimate of mean.  By assumption the sample variance is 1.0 and that tells us variance on the mean is 0.5 because 2 samples contribute to the estimate.
Now imagine that instead we sample 1.0 and 7.0.  Conditional on the model being correct this is a very unlikely situation so I would in practice assume some error had been made; that distributions were not normal; or that variance was no really 1.0.  However, in theory, if after all the checking we still believed the model then just as for the first experiment we cannot do better than 4.0 as the estimate of mean and the variance of this estimate is still 0.5.  
What changes with the observed spread is how well the model appears to fit the data.  Conditional on the model, the estimate of spread on the mean does not change at all.  
