Let $ A$ be a $2\times2$ real orthogonal matrix. Then when does $e^A$ become orthogonal as well? According to my calculations, $A$ must be skew-symmetric also and therefore there are only two possibilities. Am I correct?
Edit~~~ Ok, the two matrices are $\pmatrix{0 &1 \\ -1 &0}$ and $\pmatrix{0 & -1 \\ 1 &0}$. Thank you all for the advice below. Now, are these matrices correct? I think they are, but I would like others to check in case I am missing something.
That does indeed sound correct. (And you can probably generalize fairly easily to higher dimensions.)
As a general principle, it's worth knowing that the tangent space to the orthogonal group at $I$ consists exactly of skew-symmetric matrices; since $A \mapsto \exp A$ sends this tangent space to the orthogonal group (and indeed, ignoring speed for the moment, $t \mapsto \exp(tA)$ is a geodesic through the identity, for any skew-symmetric $A$), there's a pretty strong correspondence between the two.