Suppose that u : $\mathbb{R} \rightarrow \mathbb{R}$ is such that, whenever v : $\mathbb{R}^n \rightarrow \mathbb{R}$ is harmonic, so is u(v(x)). What can you say about u?
My first inclination is to say that properties of harmonic functions will not apply to u since it is not harmonic by definition. If that is correct, would I then only say that u is C(U) for $U \subset \mathbb{R}$? Any helpful suggestions would be appreciated.
It is easy to compute directly that $$ \nabla u(v(x)) = u'(v(x)) \nabla v(x), \\ \Delta(u(v(x))) = u'(v(x)) \Delta v(x) + u''(v(x)) (\nabla v(x))^2 $$ The first term is obviously zero, the second term is zero when at least one of its factors is zero, and for each $x$, we can choose a $v$ so that $\nabla v(x) \neq 0$. Hence $u(v(x))$ is harmonic for every harmonic $v$ if and only if $u'' \equiv 0 $.
(In the above, we assume $u$ is sufficiently smooth: otherwise the Laplacian won't make sense anyway.)