Let $A$ be an $n\times n$-matrix over a noncommutative (but associative and unital) ring. Then, $A$ is invertible if and only if its transpose $A^T$ is invertible, right?

Wrong. A counterexample (which I just learnt from Ulrich Krähmer and Blessing Bisola Oni, Pointed Hopf algebra (co)actions on rational functions, arXiv:2209.06516v3, Example 2.1.1) is the matrix $\begin{pmatrix} a&1 \\ 0&d \end{pmatrix}$ over the ring with two generators $a$ and $d$ and one relation $da=1$ (but not $ad=1$). Its inverse is $\begin{pmatrix} d&-1 \\ 1-ad&a \end{pmatrix}$, and the fact that the two matrices are mutually inverse is straightforward to verify. But its transpose $\begin{pmatrix} a&0 \\ 1&d \end{pmatrix}$ is not invertible, as its inverse would have to contain a right inverse to $a$, but then $a$ would be (two-sidedly) invertible.

Of course, this all stems from the fact that $\left(AB\right)^T$ is not $B^T A^T$ in general when the matrices $A$ and $B$ are over a noncommutative ring.