Such a set $\{v_1,v_2,\dots,v_N\}\subset\mathbb C^d$ of unit vectors having the "uniform overlap property" contains as indicated at most $d^2$ elements, because the associated rank-one projectors
$p_i=v_i\langle\,\cdot\,|v_i\rangle\in M_d(\mathbb C)$
are necessarily linear independent, and $\,\dim M_d(\mathbb C)=d^2.$
Before showing this, notice that
$$\operatorname{trace}(\,p_ip_j)
\:=\:\begin{cases}\langle v_j|v_i\rangle\,\langle v_i|v_j\rangle\:=\:C& \text{if }i\ne j\,,\\
1 & \text{if }i=j\,.\end{cases}$$
And furthermore that $0<C<1$, since $C=1$ would unmask the $v_i$ to be pairwise linear dependent, thus multiples of each other.
Consider the ansatz $\,0=\sum_{i=1}^N\mu_ip_i$, then multiply by $p_j$, and take the trace. This yields the linear equation system
$$\begin {pmatrix}
1 & C & \cdots & C \\
C & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & C\\
C & \cdots & C & 1
\end {pmatrix}\,
\begin {pmatrix} \mu_1\\ \mu_2 \\ \vdots\\ \mu_N
\end{pmatrix}\:=\:\begin {pmatrix} 0\\ 0 \\ \vdots\\ 0\end {pmatrix}$$
whose determinant equals $(1+(N-1)C)\,(1-C)^{N-1}$, cf Marc van Leeuwen's summary , hence is positive in our case. Thus the $\mu_i$ have to be zero, proving the linear independence of the $p_i$. $\;\blacksquare$
When looking at the linked paper & hopping a bit further I'd conclude with:
The existence of maximal sets with the uniform overlap property is a very much harder question, and in particular not solved for all $d$.
