I'm interested in examples of positive maps on $C^*$-algebras which are not 2-positive. I know that the transpose on the matrices is such an example, but I'm iterested in further examples and tried to construct something ( For example, I considered the map $\sigma:C_r^*(G)\to C_r^*(G)$ defined by $\sigma(\lambda_g)=\lambda_{g^-1}$ for a non-abelian discrete group $G$, but this map is 2-positive.. Or more general than that with the transpose map, I tried to construct something with the opposite $C^*$-algebra of a $C^*$-algebra, unsuccessfully).
What are other examples than the transpose map?
Consider $\mathcal A=M_2(\mathbb C)$. A linear map $\phi:M_2(\mathbb C)\to M_2(\mathbb C)$ is completely positive if and only if the $4\times4$ matrix $$\tag{1} \begin{bmatrix}\phi(E_{11})& \phi(E_{12})\\ \phi(E_{21})& \phi(E_{22})\end{bmatrix} $$ is positive (Choi's criterion). So one needs to look for maps $\phi$ that preserve positivity but that make the matrix in $(1)$ non-positive.
Being linear, $\phi$ is of the form $$ \phi(A)=\begin{bmatrix} f_{11}(A)& f_{12}(A) \\ f_{21}(A) & f_{22}(A)\end{bmatrix}. $$ for some functionals $f_{kj}$. Since $\phi$ has to be positive, we need $f_{11}\geq0$, $f_{22}\geq0$, $f_{21}(A)=\overline{f_{12}(A)}$, and $|f_{12}(A)|\leq f_{11}(A)f_{22}(A)$ for all $A\geq0$.
Every linear functional on $M_2(\mathbb C)$ is of the form $f(A)=\text{Tr}\,(XA)$ for some $X\in M_2(\mathbb C)$. Such $f$ is positive precisely when $X\geq0$.
So now we have that $$ \phi(A)=\begin{bmatrix} \text{Tr}\,(HA)& \text{Tr}\,(B^*A) \\ {\text{Tr}\,(BA)} & \text{Tr}\,(KA)\end{bmatrix}, $$ with $H,K\geq0$, and $$\tag{2}|\text{Tr}\,(BA)|^2\leq \text{Tr}\,(HA)\text{Tr}\,(KA)$$ for all $A\geq0$ is enough to guarantee that $\phi\geq0$.
Let's go back to equation $(1)$. We now have $$ \begin{bmatrix}\phi(E_{11})& \phi(E_{12})\\ \phi(E_{21})& \phi(E_{22})\end{bmatrix} =\begin{bmatrix}H_{11}&\overline{B_{11}}&H_{21}& \overline{B_{21}}\\ {B_{11}}& K_{11}& {B_{12}}& K_{21}\\ H_{12}&\overline{B_{12}}&H_{22}& \overline{B_{22}}\\ {B_{21}}& K_{12}& {B_{22}}& K_{22} &\end{bmatrix}\simeq_u\begin{bmatrix} H& B^* \\ B & K\end{bmatrix}, $$ where $\simeq_u$ denotes unitary equivalence (one needs to permute the second and third columns and rows).
Now, the problem is reduced to finding $H,B,K$ with $H,K\geq0$, satisfying $(2)$, and such that $$\tag{3}\begin{bmatrix} H& B^* \\ B& K\end{bmatrix}$$ is not positive.
One such choice is $H=E_{11}$, $K=E_{22}$, $B=E_{21}$, which produces $\phi(A)=A^T$. A slightly different choice would be $$ H=\begin{bmatrix}2&0\\0&1\end{bmatrix},\ \ K=\begin{bmatrix}1&0\\0&2\end{bmatrix},\ \ B=\begin{bmatrix}0&2\\0&0\end{bmatrix}, $$ which produces $$ \phi(A)=\begin{bmatrix}2A_{11}+A_{22}& 2A_{21} \\ 2A_{12} & A_{11}+2A_{22}\end{bmatrix}. $$ This is not particularly exciting, as $\phi(A)=2A^T+E_{12}AE_{21}+E_{21}AE_{12},$ so it still involves the transpose. But with patience one could play with many choices of $H$, $K$, and $B$.
Note, though, that the positivity in $(3)$ is decided by the $2\times 2$ block $$ \begin{bmatrix} H_{22} & \overline{ B_{12}} \\ B_{12} & K_{11}\end{bmatrix}. $$ Since $H_{22}\geq0$ and $K_{11}\geq0$, this block is always of the form $R+c\begin{bmatrix}0&1\\1&0\end{bmatrix}$ with $R\geq0$. This suggests (not a proof, though) that positive maps on $M_2(\mathbb C)$ are linear combinations of a completely positive map and a transpose-like map.