This is exercise §82.2 of Halmos 1958, 2nd ed.:
Discuss the theory of functions of a normal transformation on a [finite-dimensional] real inner product space.
In the preceding chapter, general functions on a normal transformation on a unitary (i.e. complex inner product) space are defined by applying the function to the eigenvalues.
In detail, let $A$ be normal with spectral decomposition $A=\sum_i\lambda_iP_i$. Then if is any complex-valued function defined on the eigenvalues of $A$, we define $$ f(A) = \sum_if(\lambda_i)P_i. $$
That's all well and good, but is there much else we can do in the real case? Clearly, if $A$ is real-diagonalizable (has only real eigenvalues), then we can use the same definition, except $f$ should be real-valued. However, if $A$ has complex eigenvalues, then I don't really see how to proceed without entering the complex domain.
The problem with the real case is that we lose diagonalizability. The spectral decomposition (maybe a misnomer) of a real normal matrix produces not a diagonal but a block-diagonal matrix with the real eigenvalues on the diagonal and 2-by-2 blocks of the form $$ \begin{pmatrix}a&b \\ -b&a\end{pmatrix}, $$ corresponding to eigenvalue pairs of $a\pm ib$. So we would like to make sense of $f$ applied to such a matrix, but on that point I'm lost. No matter what I think of, it would boil down to defining $f$ on the complex eigenvalues as well. If $f$ is a polynomial, we can of course just apply $f$ directly to the blocks, but that doesn't seem very interesting.
So, is there really anything more to say?
According to Qiaochu Yuan in the comments, we do indeed need $f$ to have complex numbers in the domain. In other words, there really doesn’t seem to be anything special we can salvage in the real case. Thus, I’ll resolve the question with this answer in the negative, so to speak. Anyone is welcome to add other points of view of course.
As for the intention by Halmos behind the exercise, I then suspect that he just wants the student to explore and find this dead end, and so they’ll think about why we need the complex domain. I guess.
EDIT:
A natural question that we can still ask is which functions $f$ have the property that $f(A)$ is real for any real normal transformation $A$. A necessary and sufficient condition is that $f$ commutes with conjugation, i.e. $f(\bar z) = \overline {f(z)}$. Or we could say that $f$ maps conjugate pairs to conjugate pairs.
Why? The problem reduces to two cases: $A=(a)$ and $A=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$. From here it's really a straightforward matrix calculation. Conceptually though, sufficiency is in particular due to the fact that the eigenspaces of $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ do not depend on $a$ and $b$ (they are generated by $\binom1{\pm i}$).