I recently found the following. Let $a,b\in\Bbb N$ with $24|(a+b)$ and consider the function $$f(\tau)=\eta(a\tau)\eta(b\tau).$$ We have the following symmetry relations: $$\begin{align} f(\tau+1)&=f(\tau),\tag1\\ f\left(-\tfrac{1}{ab\tau}\right)&=-i\tau\sqrt{ab}f(\tau),\tag2 \end{align}$$ which I will prove at the bottom of the post.
Considering that $\eta(\tau)$ is a modular form of weight $1/2$, is $f(\tau)$ some sort of modular form?
Just a word on notation. Throughout this question, I will use the notational conventions $\tau\in\Bbb H=\{z\in\Bbb C:\Im z>0\}$, $q:=e^{2i\pi \tau}$, and $\eta(\tau)=q^{1/24}\prod_{n\ge1}(1-q^n)=q^{1/24}(q;q)_\infty$.
Context: I was messing around with the $\eta$-function and I saw on Wikipedia that $$\eta(8\tau)\eta(16\tau)=\sum_{m,n\in\Bbb Z\\ m\le|3n|}(-1)^mq^{(2m+1)^2-32n^2},$$ which I found interesting, because the $q$-expansion $$\eta(\tau)=\sum_{n\in\Bbb Z}(-1)^nq^{(6n-1)^2/24}$$ has no integer-powers of $q$. After a little experimenting, I was able to show that the $q$-expansion for $\eta(a\tau)\eta(b\tau)$ has integer-powers of $q$ when $24|(a+b)$. So my next query was whether or not $f(\tau)$ was a modular form. Clearly the relation $(2)$ is not of the form $g(-1/\tau)=\tau^kg(\tau)$, but it is pretty similar, so I thought that maybe $f(\tau)$ was a modular form in some other sense, or for some subgroup of $\mathrm{SL}_2(\Bbb Z)$.
I also learned that $\eta$-quotients of the form $\prod_{d|n}\eta(d\tau)^{r_d}$ can be modular forms under certain restrictions on $r_d$ and $n$. Perhaps $f(\tau)$ satisfies these?
Forgive me if this question is obvious, I have only learned about this kind of thing from textbooks and Wikipedia, and I have yet to take any classes in number theory.
Proofs:
$(1)$: We can see that $$f(\tau)=\eta(a\tau)\eta(b\tau)=q^{(a+b)/24}(q^a;q^a)_\infty (q^b;q^b)_\infty=q^k(q^a;q^a)_\infty (q^b;q^b)_\infty,$$ for some $k\in\Bbb N$. Since $a,b\in\Bbb N$, clearly each product $(q^s;q^s)_\infty$, where $s=a,b$, will contribute only terms with integer-powers of $q$ to the $q$-expansion of $f$. And since $q^m$ is invariant under the transformation $\tau\mapsto\tau+1$ for integer $m$, we have that $f(\tau+1)=f(\tau)$.
$(2)$: From the relation $\eta(-\tfrac1{\tau})=\sqrt{-i\tau}\eta(\tau)$ it is easy to prove $(2)$.