Time reverse transformation of 4-potential and its relation to Lorentz transformation

Until now, I thought electromagnetic potential $A^{mu}(x)$ transform like $x^{mu}$ under the Lorentz transformation:
$$A^{mu}(x)=Lambda^{mu}_{ nu}A^{nu}(x).$$
But according to time reversal symmetry of governing equations, $A^{mu}(x)=(A^{0}(x), A^{j}(x))$ transform under the time inverse transformation:
$$A’^{mu}(x’)=(A^{0}(x), -A^{j}(x))neq (-A^{0}(x), A^{j}(x))= Lambda^{ mu}_{T nu}A^{nu}(x）$$ (where $ Lambda_{T}$ is the time inverse transformation and first equal follows from $B’(x’)=-B(x)$ & $E’(x’)=E(x)$ under the time inverse). I feel this is inconsistent.

In addition to this, according to the Smith’s linked PDF (http://www.physics.usu.edu/Wheeler/TopicPlugins/SpecialRelativityMaster.pdf#page17), usual time reversal transformation $T$ ( which replace $t$ by $-t$) is different from time reversal in Lorentz transformation: $Lambda_{T nu}^{ mu}=diag(-1,1,1,1). $ This is because Sometimes $T$ does not keep Lorentz norm (e.g. $(x^{mu}+p^{mu})(x_{mu}+p_{mu})$), and it does not act all 4-vectors in same way (e.g. $p’^{mu}=(E, -{ bf p}) neq Lambda_{T nu}^{ mu}p^{nu})$. I guess this is true, but why is $Lambda_{T nu}^{ mu}=diag(-1,1,1,1)$ called as time reversal? I feel this name is very misleading.

Can somebody explain this? I would misunderstand 4-vectors and concepts of covariant and contravariant.