If p is prime, then there exists $d < p$ such that $d^{p-1} \equiv 1 \pmod p$, or equivalently, $p|d^{p-1}-1$?
Also, if possible, prove that at the same time,$d^{p-1} \not\equiv 1 \pmod p$ must not hold for any $v<p-1$.
I am not sure if it is true. But if the function group $\operatorname{Aut}(\mathbb{Z}_p)$ of the automorphism of $\mathbb{Z}_p$ under function composition is homomorphic to the group $\mathbb{Z}_{p-1}$, where $\mathbb{Z}_p$ is the group of integers modulo $p$ under addition, then my assumptions above should be true.
Your problem is equivalent to proving that there is a primitive root modulo $p$ for prime integers $p$. This is a well-known fact and a quick search on the site yields this answer as nice proof of it.
In the same time this proves that $U(p) \cong \operatorname{Aut}(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$, which is hat you wanted. Taking any generator of $U(p)$, the group of units modulo $p$ will satisfy the condition you want.