Is it possible to have 2 different but equal size real number sets that have the same mean and standard deviation? By inspection I notice that


*

*Shifting does not change the standard deviation but change mean.
{1,3,4} has the same standard deviation as {11,13,14} for example.

*Sets with the same (or reversed) sequence of adjacent difference have the same standard deviation. For example, {1,3,4}, {0,2,3}, {0,1,3} have the same standard deviation. But the means are different.
My conjecture: There are no two distinct sets with the same length, mean and standard deviation. 
Question
Is it possible to have 2 different but equal size real number sets that have the same mean and standard deviation?

You seem to like the set $\{1,3,4\}$.  Here are six more three element sets having the same mean and standard deviation as your given set.  \begin{align*}
 &\frac{202}{171} & &\frac{1}{171} \left(583+\sqrt{19842}\right) & &\frac{1}{171}
   \left(583-\sqrt{19842}\right) \\
 &\frac{688}{171} & &\frac{1}{171} \left(340+\sqrt{27861}\right) & &\frac{1}{171}
   \left(340-\sqrt{27861}\right) \\
 &\frac{544}{171} & &\frac{1}{171} \left(412-\sqrt{62421}\right) & &\frac{1}{171}
   \left(412+\sqrt{62421}\right) \\
 &\frac{32}{19} & &\frac{1}{19} \left(60-\sqrt{581}\right) & &\frac{1}{19}
   \left(60+\sqrt{581}\right) \\
 &\frac{27}{19} & &\frac{1}{38} \left(125-\sqrt{1689}\right) & &\frac{1}{38}
   \left(125+\sqrt{1689}\right) \\
 &-\frac{2}{3} \left(-4+\sqrt{7}\right) & &\frac{1}{6} \left(16+2 \sqrt{7}\right) &
   &\frac{1}{3} \left(8+\sqrt{7}\right)
\end{align*}
and, more generally, let $(x,y)$ be any point on the ellipse given by the equation 
$$  x^2 + xy + y^2 -8x -8y + 19 = 0  $$
and set $z = 8 - x - y$.  This triple of values has the same mean and standard deviation as does $\{1,3,4\}$.  (This is found by eliminating $z$ from the system mean$(x,y,z) = {}$mean$(1,3,4)$ and stddev$(x,y,z) = {}$stddev$(1,3,4)$, i.e., $x+y+z = 8, x^2 + y^2 + z^2 - xy - xz - yz = 7$.)
