Consider a real scalar field, $$\phi(x)=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}\right),$$ here the function $v_{\vec{k}}(t)$ satisfies the normalization condition, $$v_{\vec{k}}\dot{v}^*_{\vec{k}}-v^*_{\vec{k}}\dot{v}_{\vec{k}}=i.$$ Now, changing the variable $\vec{k}\rightarrow-\vec{k}$ in the fourier expansion we have,$$\phi(x)=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}\right).$$ Comparing the coeeficeints of $e^{i\vec{k}\cdot\vec{x}}$ and its complex conjugate we have, $$a_{\vec{k}}v_{\vec{k}}=a^\dagger_{-\vec{k}}v^*_{-\vec{k}}.$$ From this can it be said that $v_{\vec{k}}=v^*_{-\vec{k}}$ and $a_{\vec{k}}=a^\dagger_{-\vec{k}}$ because both the operators and functions should be identified separately? If yes then why and if not then why?
Edit: As a comment mentioned that it is not clear what two expressions, I am comparing, I shall write it explicitly, $$\begin{align} \phi(x)&=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}\right)\\&=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}\right)\end{align}$$. Thus we can write, $$a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}=a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}.$$ Notice that this equality holds for each $k$ mode separately and there is no integration anymore. Now we compare the coefficients of the modes from LHS and RHS of the equation.