Let $A$ be a SPD matrix. Suppose diagonal $(A)_{ii}$ equals to its eigenvalue $\lambda_i$. Must $A$ be a diagonal matrix?

$\newcommand\La\Lambda$As noted in my comment, it is not necessary that $U=I$.

However, given the conditions $A=U\Lambda U^{\top}$ and $\operatorname{diag}(A)=\operatorname{diag}(\Lambda)$, it is necessary that $A$ be a diagonal matrix (so that $A=\Lambda$). Indeed, by Hadamard's inequality for psd matrices, $$\det A\le\prod_i A_{ii},$$ with the equality only if $A$ is diagonal. Here $M_{ij}$ stands for the $(i,j)$-entry of a matrix $M$. It remains to note that $$\det A=\det\Lambda=\prod_i \Lambda_{ii}=\prod_i A_{ii}.$$

Also, if the $\La_{ii}$'s are pairwise distinct, then $U$ is a diagonal matrix such that $U^2=I$. Indeed, then without loss of generality $n\ge2$ and $\La_{11}>\cdots>\La_{nn}$, if $A$ is $n\times n$. The conditions $A=U\Lambda U^{\top}$ and $\operatorname{diag}(A)=\operatorname{diag}(\Lambda)$ imply $$\La_{ii}=\sum_j U_{ij}^2\La_{jj}$$ for all $i$, so that we get the contradiction $$\La_{11}=\sum_j U_{1j}^2\La_{jj}<\max_j\La_{jj}=\La_{11}$$ unless $U_{11}^2=1$ and hence $U_{1j}=0=U_{j1}$ for $j\ge2$; this follows because $U$ is orthogonal. So, by induction on $n$, we get $U_{ij}^2=1(i=j)$ for all $i,j$ -- that is, $U$ is a diagonal matrix such that $U^2=I$. On the other hand, obviously, if $U$ is a diagonal matrix such that $U^2=I$, then $A=U\Lambda U^{\top}=\La$.