Since $\det(A) \neq 0$, $A$ is nonsingular and all diagonals of $A$ must equal one. Next, we know the relationship between the determinant and trace is
\begin{align*}
\det(A) &= \exp(tr(\ln(A))) \\
&= \exp(0) \\
&= \exp(\sum_i \ln(\lambda_i)) \\
\end{align*}. Step 2 holds because all diagonals of $A$ are one, so their log is zero, and step 3 because natural logs are analytic. This suggests $$\sum_i \ln(\lambda_i) = 0$$, or, $$\prod_i \lambda_i = 1$$.

Then, because all diagonals of $A$ are one, $tr(A) = \sum_i \lambda_i = n$ must be true. Hence, the AM-GM inequality states $$ \left[ \prod_{i} \lambda_i \right]^{\frac{1}{n}} \leq \frac{1}{n} \sum_{i} \lambda_{i}$$, with equality holding *only* when $\lambda_i = \lambda_j$ $\forall i,j$. Thus, $$ 1^{\frac{1}{n}} = \frac{1}{n} \cdot n$$, and all eigenvalues of $A$ must be equal. Since $\prod_i \lambda_i = 1$, all eigenvalues must be one.