In Diamond and Shurman's A First Course in Modular Forms, the authors define meromorphy at infinity as follows:
Let $f$ be a meromorphic function on the upper half plane that is weakly modular with respect to a congruence subgroup $\Gamma$, and has no poles with imaginary part greater than some $c$. Let $h$ be the smallest positive integer such that $\begin{pmatrix} 1 & h \\\ 0 & 1 \end{pmatrix} \in \Gamma$. Then $f$ has a Laurent series on the punctured disk around $0$ corresponding to the region $\{\tau \in \mathcal{H} : \operatorname{Im}(\tau) > c\}$ given by $$f(\tau) = \sum_{n=-\infty}^\infty a_n q^{n/h},$$ where $q(\tau) = e^{2\pi i (\tau)}$. $f$ is then meromorphic at $\infty$ if this Laurent series is truncated to the left, and the order is the minimal $m$ such that $a_m \neq 0$.
I'm wondering about the dependence in the definition of the order of $f$ at infinity on choosing the smallest $h$ such that $\begin{pmatrix} 1 & h \\\ 0 & 1 \end{pmatrix} \in \Gamma$. For any $n$ such that the corresponding matrix is in $\Gamma$, we have that $f$ is $n$-periodic. Will the order at infinity change if we don't choose the smallest possible $n$?
I realize this is likely easy to answer by computing the $q$-expansions directly and seeing how they change, but honestly I have rarely needed to work with $q$-expansions explicitly, so I don't feel like I have the tools to see the answer myself.
Assuming $f$ is not constant, any other integer $n$ such that $f$ is $n$-periodic will be a multiple of $h$ (which I'll use to denote the smallest possible period).
Denote $q_N=\exp(2\pi i \tau/N)$. Then a relation between a $q_h$ expansion and a $q_{mh}$-expansion follows from the simple observation that $q_{mh}^m=q_h$. Using this and the fact that Laurent series are unique, we can easily prove that changing $h$ by a factor multiplies the order of vanishing at infinity by the same factor.