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?

The example by auscrypt settles the question, but maybe it's worth mentioning why this should be obvious by considering degrees of freedom.  
Mean and standard deviation are two quantities.  A collection of $m$ real numbers has $m$ degrees of freedom.  Specifying the mean and standard deviation removes two degrees of freedom, leaving $m-2$.  So as long as  $m > 2$, there should still be lots of room to have
different sets with the same mean and standard deviation.
EDIT: This was intended as a heuristic, rather than a proof, but rigorous arguments can be made.  For example, suppose $A$ and $B$ are two $m$-tuples with
the same mean, such that $\sigma(A) > \sigma(B)$ (where $\sigma$ denotes standard deviation), $C$ and $D$ two $m$-tuples with the same mean such that $\sigma(C) < \sigma(D)$.  Then for any $t$, $t A + (1-t) C$ and $t B + (1-t) D$ have the same
mean, and (by the Intermediate Value Theorem) there exists $t \in [0,1]$ such that they have the same standard deviation.  If $A-B$ and $C-D$ are linearly independent these will not be the same.
