Deterministic case
The map sends
$$
|+\rangle|+\rangle\mapsto|+\rangle|+\rangle\tag1
$$
and
$$
\begin{align}
|0\rangle|+\rangle\mapsto&|0\rangle|-\rangle\tag2\\
|1\rangle|+\rangle\mapsto&-|1\rangle|-\rangle\tag3
\end{align}
$$
so it fails to be linear (let alone unitary) and therefore cannot be realized as a quantum circuit.
Probabilistic case
Let $\mathcal{E}$ be a completely positive, but not necessarily trace-preserving linear map that implements the desired probabilistic protocol. Then, similarly to $(1)$-$(3)$, we have
$$
\begin{align}
\mathcal{E}(|+\rangle\langle+|\otimes|+\rangle\langle+|)&=p_1|+\rangle\langle+|\otimes|+\rangle\langle+|\tag4\\
\mathcal{E}(|-\rangle\langle-|\otimes|+\rangle\langle+|)&=p_2|-\rangle\langle-|\otimes|+\rangle\langle+|\tag5
\end{align}
$$
and
$$
\begin{align}
\mathcal{E}(|0\rangle\langle 0|\otimes|+\rangle\langle+|)&=p_3|0\rangle\langle 0|\otimes|-\rangle\langle-|\tag6\\
\mathcal{E}(|1\rangle\langle 1|\otimes|+\rangle\langle+|)&=p_4|1\rangle\langle 1|\otimes|-\rangle\langle-|\tag7
\end{align}
$$
for some $p_1,p_2,p_3,p_4\in[0,1]$. But the maximally mixed state $$
I=\frac12|0\rangle\langle 0|+\frac12|1\rangle\langle 1|=\frac12|+\rangle\langle +|+\frac12|-\rangle\langle -|\tag8
$$
so $\mathcal{E}(\frac{I}{2}\otimes|+\rangle\langle+|)$ is simultaneously proportional to $\rho_1\otimes|+\rangle\langle+|$ and to $\rho_2\otimes|-\rangle\langle-|$ for some states $\rho_1$ and $\rho_2$. This is possible only if $\mathcal{E}\left(\frac{I}{2}\otimes|+\rangle\langle+|\right)=0$. Similarly, $\mathcal{E}\left(\frac{I}{2}\otimes|-\rangle\langle-|\right)=0$. Consequently, $\mathcal{E}\left(\frac{I}{4}\right)=0$. But $\mathcal{E}$ is completely positive, so $\mathcal{E}(\rho)=0$ for all $\rho$.
Given an input state $\rho$, the probability that the process described by $\mathcal{E}$ occurs is $\mathrm{tr}(\mathcal{E}(\rho))$. But $\mathcal{E}(\rho)=0$, so this probability is zero.