Spinor chiral transformation by $\psi \to \gamma^5 \psi$ Let $\psi$ be a spinor. Let $\gamma^0,\gamma^1, \gamma^2, \gamma^3$ be the usual gamma matrices and the fifth $\gamma^5 : = i\gamma^0\gamma^1\gamma^2\gamma^3.$
Then if we define $\psi \to \psi' := \gamma^5 \psi$ the chiral transformation I want to show what it will be the transformations for $\overline{\psi}\psi$ were $\overline{\psi} := \psi^\dagger \gamma^0$. 

My question is how can I understand this transformation under 
  $\overline{\psi}\psi$? 

In the problem sheet is not at all clear, to me, that we could just use $(\psi_1\psi_2)'=\psi_1'\psi_2'$ and then perform the transformation. And how $\overline{\psi}$ should be transformed? It is also not defined. One possibility is that 
$$\overline{\psi}' = (\psi^\dagger\gamma^0)'= \psi'^\dagger\gamma^0 = \psi^\dagger\gamma^5\gamma^0 = - \overline{\psi}\gamma^5$$
Were I used that $\{\gamma^5,\gamma^\mu\}=0$, for all $\mu=0,1,2,3$. Other possibility is just
$$\overline{\psi}' := \overline{\psi}\gamma^5$$
were I swich for a left operation of the gamma because $\overline{\psi}$ is a row spinor (is this correct in some sense?) and also, if we speak only with spinors I could just don't change at all for the $\overline{\psi}$ and just set $\overline{\psi}' = \gamma^5\overline{\psi}$. 
Thanks in advance. This is more a question on trying to understand what is asked and how this transformation is applied. 
 A: You may put it in this way.
Let $\psi'=\mathcal{C}[\psi]$ be a general transformation of $\psi$, which, in your chiral transformation case, reads
$$
\mathcal{C}[\psi]=\gamma^5\psi.
$$
Let $\mathcal{L}[\psi]$ be a general quantity to be focused on, which, in your question, is defined as
$$
\mathcal{L}[\psi]=\bar{\psi}\psi=\psi^{\dagger}\gamma^0\psi.
$$
Here I use bracket "$[\cdot]$" instead of parenthesis "$(\cdot)$", as $\mathcal{C}$ and $\mathcal{L}$ could be general mappings rather than usual functions. For example, in quantum electrodynamics, $\mathcal{L}$ could go by
$$
\mathcal{L}[\psi]=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}\psi-\psi\right),
$$
which involves not only $\psi$ itself, but its derivatives as well.
With these notations, the transform of $\mathcal{L}$ under the transformation $\mathcal{C}$ is defined as
$$
\mathcal{L}'=\mathcal{L}\circ\mathcal{C},
$$
namely
$$
\mathcal{L}'[\psi]:=\mathcal{L}[\psi']=\mathcal{L}[\mathcal{C}[\psi]].
$$
So in your question, the transform of $\psi^{\dagger}\gamma^0\psi$ under chiral transformation should be
$$
\left(\psi^{\dagger}\gamma^0\psi\right)'=\mathcal{L}'[\psi]=\mathcal{L}[\psi']=\mathcal{L}[\gamma^5\psi]=\left(\gamma^5\psi\right)^{\dagger}\gamma^0\left(\gamma^5\psi\right)=\psi^{\dagger}\left(\gamma^5\right)^{\dagger}\gamma^0\gamma^5\psi=-\psi^{\dagger}\gamma^0\psi,
$$
where the second step from the last is due to
$$
\left(\gamma^5\psi\right)^{\dagger}=\psi^{\dagger}\left(\gamma^5\right)^{\dagger},
$$
while the last step is because
$$
\left(\gamma^5\right)^{\dagger}=\gamma^5\quad\text{and}\quad\gamma^5\gamma^0\gamma^5=-\gamma^0.
$$
In general, as $\mathcal{C}$ transforms $\psi$ into $\psi'$, you need to replace every $\psi$ that appears in $\mathcal{L}$ by $\psi'$, including those in $\bar{\psi}$, $\psi^{\dagger}$, $\partial_{\mu}\psi$, etc. The result that you obtain is then the correct one for $\mathcal{L}'$, the transform of $\mathcal{L}$ under the transformation $\mathcal{C}$.
Hope this would be helpful for you.
