The other possibility is that $f(z)=e^z$. Explicit computation gives
$$\frac{d}{dr}f^\#(re^{i\phi})=2e^{r\cos\phi}\cos\phi(1-e^{2r\cos\phi}),$$
which is $\leq 0$ everywhere.

On the other hand, the exponential is the only exception.

Indeed, if $f^\#$ is decreasing on each radius, then $f^\#$ has no zeros,
this implies that $f'$ has no zeros. On the other hand, $f^\#(z)\leq f^\#(0)$ must be bounded, and a theorem of Clunie and Hayman implies that $f$ has at most exponential type. Then $f'$ has at most exponential type, and since it has no zeros, it must be $e^{az+b}$ by the Hadamard factorization theorem.

Refs. The original paper of Clunie and Hayman is

MR0192055 
Clunie, J.; Hayman, W. K.
The spherical derivative of integral and meromorphic functions.
Comment. Math. Helv. 40 (1966), 117–148.

Their proof was much simplified (and generalized) in

MR2869124 
Barrett, Matthew; Eremenko, Alexandre 
Generalization of a theorem of Clunie and Hayman. 
Proc. Amer. Math. Soc. 140 (2012), no. 4, 1397–1402.