Let us say that we are given a DFT matrix $\Omega_n$ with $2$ signals $z, w \in \mathbb{C^n}$, such that $\hat{z}=\Omega_n z, \hat{w}=\Omega_n w$ are the discrete Fourier Transformation from $z$ and $w$. If the columns of $\Omega_n$ are normalised, then can we say that $\widehat{z_{n-1}}$ describes the share of the highest frequencies of $z$?
Can we then use this to say that, the convolution of $a$ and $a$ corresponds to the elemental multiplication of the transformed signals: $z ∗ w = \hat{z}⋅\hat{w}$?
As discussed in the comments, the convolution property of the DFT means that $\widehat{z*w} = \hat z \cdot \hat w$. The statement $z*w = \hat z \cdot \hat w$ does not hold in general.
Regarding $\hat z_{n-1}$: because of aliasing, calling anything a "high frequency component" is a bit tricky. But if we think of $e^{\frac{2 \pi i (N-1)}{N}n }$ as being the signal of highest frequency, then yes your interpretation of $\hat z_{n-1}$ makes sense. Keep in mind, however, that we could have written the same signal as $e^{\frac{-2 \pi i }{N}n }$, which we would normally think of as having a low frequency.