Suppose A is a symmetric $n\times n$ matrix. $A$ is diagonally dominant (dd) if:
$$ |A_{ii}| \ge \sum_{j=0, j\neq i}^n |A_{ij}| \ \ \forall i=1,2,\cdots ,n $$
Is $U^TAU$ dd for every dd $A$ and orthogonal $U$?
If we limit matrices $U$ to be permutation matrices instead of orthogonal matrices then $U^TAU$ is dd.
Proof: suppose $U=U_1 U_2 \cdots U_n$ where each $U_i$ is a transposition matrix. Then
$$U^TAU=U_n^T\cdots U_1^TAU_1^T\cdots U_n^T.$$
Suppose $\bar{A}=U_1^TAU_1^T$. We permute the rows and the columns $p$ and $q$ of the matrix $A$ by the conjugation by $U_1$. Then
$$ |A_{qq}| \ge \sum_{j=0, j\neq q}^n |A_{qj}|$$ implies $$ |\bar{A}_{pp}| \ge \sum_{j=0, j\neq q}^n |A_{qj}|= \sum_{j=0, j\neq q}^n |\bar{A}_{pj}|$$ and $$ |A_{pp}| \ge \sum_{j=0, j\neq p}^n |A_{pj}|$$ implies $$ |\bar{A}_{qq}| \ge \sum_{j=0, j\neq p}^n |A_{pj}|= \sum_{j=0, j\neq q}^n |\bar{A}_{qj}|$$
Hence, $\bar{A}$ is dd. By continuing this procedure, we see $U^TAU$ is dd.
However, I cannot prove for orthogonal matrices U.
It should be easy to cook up a counterexample. Let $A=\operatorname{diag}(1,0,\ldots,0)$. Then $U^TAU=uu^T$ where $u$ is the first column of $U^T$. Now, whenever $0<|u_i|<|u_k|$ for some $i$ and $k$, the matrix $B=U^TAU=uu^T$ is not diagonally dominant, because $b_{ii}=u_i^2<|u_iu_j|=|b_{ik}|\le\sum_{j\ne i}|b_{ik}|$.
For instance, let $U^T=\pmatrix{\frac{\sqrt{3}}{2}&\frac{-1}{2}\\ \frac{1}{2}&\frac{\sqrt{3}}{2}}$ and $A=\pmatrix{1&0\\ 0&0}$. Then $U^TAU=\pmatrix{\frac{3}{4}&\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4}&\frac{1}{4}}$ is not diagonally dominant on the second row.
Since the elements of $U^TAU$ are continuous in the elements of $A$, if you perturb $A$ slightly, such as by taking $A=\pmatrix{1&0\\ 0&\epsilon}$ or $\pmatrix{1&\epsilon\\ \epsilon&\epsilon}$ for some small $\epsilon>0$, you may obtain also a counterexample with the same $U$ but a strictly diagonally dominant $A$.