Assuming that $f(z)$ is a non-zero weight-$k$ automorphic form (weakly holomorphic modular form), is it possible for $f$ to satisfy
$$f|_kS=-f,\quad f|_kT=f,$$
where $f|_kS=z^{-k}f(-1/z)$ and $f|_kT=f(z+1)$.
If it is possible, please provide an example; if not, please provide a proof.
My conjecture is that there is no such $f$, and my attempt is as follows. Assuming $f|_kS=-f, f|_kT=f$, then
$$f|_k(-I)=f,\quad f|_k(T^2)=f,\quad f|_k(ST^2S^{-1})=f.$$
Since $\Gamma(2)=\langle -I,T^2,ST^2S^{-1} \rangle$, we have $f\in M^!_k(\Gamma(2))$(space of weakly holomorphic modular forms, spanned by theta functions), so $\Delta^n f\in M_t(\Gamma(2))$ for some $n, t$, where $\Delta$ is the normalized weight $12$ cusp form of level $1$. However, when I tried small values of $t$ in $M_t(\Gamma(2))$, I found that there is no such $f$.
All comments are welcome!
If you had such a form, then sending $\gamma$ to the constant by which $\gamma$ acts on $f$ would define a group homomorphism $\operatorname{SL}_2(\mathbb{Z}) \to \{\pm 1\}$ sending $T$ to $+1$ and $S$ to $-1$. (Conversely, if such a homomorphism existed then it's guaranteed that there would be some $f$ with this transformation law, although that's much deeper.)
Now, the elements $S$ and $T$ satisfy some relations: we have $S^2 = (ST)^3$, and $S^4 = 1$. Can you see why these relations forbid the existence of this homomorphism?