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

$-2,-1,3$ and $-3,1,2$ both have a mean of $0$ and a standard deviation of $\sqrt\frac{14}{3}$.

Yes. Two sets of numbers has the same mean and the same SD iff their sum and the sum of their squares match.

The set $\{1,2,3\}$ has sum $6$ and squares' sum $14$. The set $\{x,y,z\}$ the same mean and SD iff $$\begin{cases}x+y+z=6\\x^2+y^2+z^2=14\end{cases}$$ This is the intersection of a spherical surface and a sectioning plane, that has certainly infinitely many points.

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.