Consider the map $L(S) = USU^T + V$, where $U, S, V$ assumed compatible, real matrices.
If $V \succ 0$, $U$ has all eigenvalues in the interior of the complex unit disk, then it is known that $L$ has a unique fixed point $S^\ast \succ 0$.
Is it also the case that $L$ is contractive in say the operator norm, so that repeated applications of $L$ attract us to $S^\ast$?
Here $U,V,S\in M_n(\mathbb{R})$ are any matrices and $\Delta$ is the complex unit disk.
For every operator norm (induced norm), one has
$||L(S_1-S_2)||=||U(S_1-S_2)U^T||=||(U\otimes U)(S_1-S_2)||$ -if we stack the matrices row by row- cf.
https://en.wikipedia.org/wiki/Kronecker_product
$||L(S_1-S_2)||\leq ||S_1-S_2||||U\otimes U||$.
If $spectrum(U)=(\lambda_i)_i\subset \Delta$, then $spectrum(U\otimes U)=(\lambda_i\lambda_j)_{i,j}\subset \Delta$, that implies $\rho(U\otimes U)=\rho(U)^2<1$.
For small $\epsilon >0$, there is an operator norm $N$ s.t. $N(U\otimes U)<\rho(A)^2+\epsilon<1$. Finally, $L$ is a contraction mapping (for $N$) and, when $k\rightarrow +\infty$, $L^k(S)$ tends to $S^*$ with $N(L^k(S)-S^*)=O((\rho(A)^2+\epsilon)^k)$.