Is a unary language regular iff its exponent is a linear function?
Linear is close, but the technical term you're looking for is semilinear:, that is, a finite union of linear sets.

Half of the proof of this is a corollary of Parikh's Theorem, which says that any context-free language has a semi-linear Parikh map (that is, the set of vectors containing the occurrences of each letter in the alphabet).

For a unary language, the parikh map of the language is the language itself (i.e. each word is uniquely identified by how many letters it has), so every unary regular language is semi-linear.

The other half of the proof is showing that you can construct a regular language containing every unary semi-linear set. This takes a bit of work, but it's not too hard if you use regular expressions:


$a^k$ recognizes the language $\{ k \}$
$(a^k)^*$ recognizes $\{xk \mid x \in \mathbb{N}_0 \}$
$R_1 R_2$ recognizes $S_1 + S_2$ if $R_1$ recognizes $S_1$ and $R_2$ recognizes $S_2$, where $+$ here is element-wise addition
$R_1 | R_2$ recognizes $S_1 \cup S_2$ if $R_1$ recognizes $S_1$ and $R_2$ recognizes $S_2$, where $|$ here is regular-expression union.