To what extent can the measures of the properties of data sets restrict the possibility of what they can be? Despite the trivial cases, such as 1,1 and 1,1,1, can there be two different sets of data that have the same central tendency and dispersion? 
 A: Yes. For example: the data 1, 2, 3, 6 have mean 3 and SD 2.16.
And so do the data 0, 3, 4, 5. One dataset is the mirror
image of the other. 
These two datsets have same same absolute deviations
$|X_i - \bar X|$ from the mean $\bar X = 3,$ but in the opposite order.
You can use this idea to make examples of larger sizes.
That's just for matching the means and the SDs. But you used generic terms 'central 
tenency' and 'dispersion'.  Here is a modification of the previous example in which both datasets have the same mean, median, and (not shown in Minitab printouts) mode.
Also the same variance, SD, IQR, range, and (not shown) mean absolute deviation.
 Data Display 

  Row  x  y
    1  1  0
    2  2  3
    3  3  3
    4  3  3
    5  3  4
    6  6  5

 Descriptive Statistics: x, y 

 Variable  N   Mean  SE Mean  StDev  Variance  Median  Range    IQR
 x         6  3.000    0.683  1.673     2.800   3.000  5.000  2.000
 y         6  3.000    0.683  1.673     2.800   3.000  5.000  2.000

For populations instead of datasets, the random variables
$X \sim Norm(\mu=3, \sigma=3)$ and $Y \sim Exp(mean=3)$ have $E(X) = E(Y) = SD(X) = SD(Y) = 3.$ Large samples from these distributions will tend to have sample means and SDs that match the corresponding population values. Graphs of their densities are
shown below.

Here are sample means and SDs for random samples of size $n = 100,000$ from each
distribution (simulated in R).
 x = rnorm(10^5, 3, 3); mean(x); sd(x)
 ## 2.996163  # Approx E(X) = 3
 ## 2.99926   # Approx SD(X) = 3
 y = rexp(10^5, 1/3); mean(y); sd(y)  # rate 1/3 implies mean 3 & SD 3
 ## 3.006326  # Approx E(Y) = 3
 ## 3.007839  # approx SD(Y) = 3

A: "Sets of data" perhaps implies finite lists of numbers.
Suppose $x_1,\ldots,x_n$ and $y_1,\ldots,y_m$ are two such lists.
Let $\bar x$ and $\bar y$ be the respective sample means and let $s_x$ and $s_y$ be the sample standard deviations.
Then $\left( \dfrac{x_i-\bar x}{s_x} : i = 1,\ldots,n \right)$ and $\left( \dfrac{y_i-\bar y}{s_y} : i = 1,\ldots,m \right)$ have the same "central tendency" (i.e. sample mean), which is $0$, and the same dispersion, i.e. the same standard deviation, which $\text{is } 1$.
Now suppose we want some other mean, say $73$, and some other standard deviation, say $14$.  Then we can use $\left( 73 + 14 \dfrac{x_i-\bar x}{s_x} : i = 1,\ldots,n \right)$ and $\left( 73 + 14 \dfrac{y_i-\bar y}{s_y} : i = 1,\ldots,m \right)$.
