$\left<\psi\right|$ is not a state of a quantum system, it is a linear functional that takes a quantum state and returns a scalar. In terms of basic Linear Algebra, it is a row vector rather than a column vector, and the conjugate transpose of $\left|\psi\right>$. So $\left<\phi|\psi\right>$ is a scalar (inner product) while $\left|\psi\right>\left<\phi\right|$ is a matrix (outer product).
You may have wanted instead to convert $\left|\psi\right>$ to another ket that has all of its probability amplitudes on a basis complex conjugated, without transposing the vector, e.g. turning $\left|\psi\right> = \sum_{x = 0}^Na_x\left|x\right>$ to $\sum_{x = 0}^N \bar a_x\left|x\right>$. There is no single unitary transformation that can take any arbitrary state and complex conjugate this way since unitary transformations by their definition preserve complex inner product and if two different states go through this transformation their new inner product together would be conjugated compared to prior, so any inner products that aren't real numbers wouldn't be preserved.
If, however, you know in advance what the probability amplitudes of $\left|\psi \right>$ are in the computational basis, then a unitary transform that will complex conjugate at least that quantum state is $C_{jk} = \delta_{jk}e^{-2i \phi_j}$ where $\phi_j$ is the argument of $a_j$ in $\left|\psi\right>$. If you want to convert the probability amplitudes of $U\left|p_0\right>$ to their conjugates given knowledge of $U$ and $\left|p_0\right>$, then such a transformation is $\bar U C\left|p_0\right>$, where $C$ conjugates the amplitudes of $\left|p_0\right>$.