Let $X$ be $\Bbb R^d$-valued random variable with $\varphi_{X}$ is characteristic function and $\Bbb P_{X}$ its distribution. I want to use the change-of-variables formula for pushforward measures ($\ast$) to show that $\Bbb P_{X} = \Bbb P_{-X} \Rightarrow \varphi_{X} = \varphi_{-X}$.
$\varphi_{-X}(t) = \varphi_{X}(-t) = \int_{\Bbb R^d} e^{i \langle -t,s \rangle} d\Bbb P_{X}[ds] = \int_{\Bbb R^d} \phi \circ A d\Bbb P_{X}[ds]$ where $\phi: \Bbb R^d \to \Bbb R, s \mapsto e^{i \langle t,-s \rangle}$ and $A = -I_d$.
Then, $\int_{\Bbb R^d} \phi \circ A d\Bbb P_{X}[ds] = \int_{\Bbb R^d} \phi \circ A d\Bbb P_{X}[ds \circ A^{-1}]$ by $(\ast)$.
Using that $A^{-1} = A$ and $\Bbb P_{X}(-A) = \Bbb P_{X}(A)$ for all $A \in \mathcal B^d$ gives us the result.
If $\mathbb P_X=\mathbb P_{-X}$ then $Ef(X)=Ef(-X)$ for any bounded measurable function $f$. Take $f(x)=e^{itx}$ to get $\phi_X=\phi_{-X}$. Your argument is correct but unnecessary.