Is this operator $A = \pmatrix{1&1\\0&1}$ self-adjoint? Is this operator
$$A = \pmatrix{1&1\\0&1}$$
self-adjoint? I think not, because
$$\pmatrix{1&1\\0&1}^T\neq A$$
where $T$ is the transposition of the matrix. What do you all think?

No, the operator is not self adjoint.  Your reasoning is correct.
Note that the definition of the adjoint of an operator is not simply that the adjoint of $A$ is $A^T$.  However, in the context of real matrices taken with respect to the standard dot product or with respect to the standard dual basis, this is what "adjoint" comes to mean.  An operator is "self-adjoint" if it is equal to its adjoint, which for real matrices means that $A = A^T$.
