$SU(2)$ symmetry of $\mathcal{L}=\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - \Phi^{\dagger}M\Phi$ I'm considering a Lagrangian of two complex scalar field:
$$\mathcal{L}=\partial_{\mu}\phi_1^{*}\partial^{\mu}\phi_1-m_1^2\phi_1^{*}\phi_1+\partial_{\mu}\phi_2^{*}\partial^{\mu}\phi_2-m_2^2\phi_2^{*}\phi_2$$
It can be written in a doublet:
$$\Phi_1 =\begin{pmatrix}
      \phi_1\\
      \phi_2
    \end{pmatrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,
\Phi_1^{\dagger} = \begin{pmatrix}
      \phi_1^\dagger &
      \phi_2^\dagger
    \end{pmatrix}
\,\,\,\,\,\,\,\,\,\,\,\,\,
M = \begin{pmatrix}
      m_1 & 0 \\
      0 & m_2 
    \end{pmatrix}
$$
$$\mathcal{L}=\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - \Phi^{\dagger}M\Phi$$
It has an internal global symmetry $SU(2)$:
$$
\begin{cases}
\Phi^{'}= e^{\frac{i}{2}\vec\alpha \cdot \vec\sigma} \Phi\\
\Phi^{'\dagger} = \Phi^{\dagger}e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} \end{cases} $$
I'd like to check it explicitly but I'm stuck:
$$\mathcal{L^{'}}=\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi - \Phi^{\dagger}M\Phi$$
$$\mathcal{L^{'}}=\begin{pmatrix}
      \partial_{\mu}\phi_1^{*} &
      \partial_{\mu}\phi_2^{*}
    \end{pmatrix} e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma}
    e^{+\frac{i}{2}\vec\alpha \cdot \vec\sigma}
 \begin{pmatrix}
      \partial^{\mu}\phi_1 \\
      \partial^{\mu}\phi_2 
    \end{pmatrix}
-
\begin{pmatrix}
      \phi_1^{*} &
      \phi_2^{*}
    \end{pmatrix}
e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma}
 \begin{pmatrix}
      m_1 & 0 \\
      0 & m_2 
    \end{pmatrix} 
e^{+\frac{i}{2}\vec\alpha \cdot \vec\sigma}
\begin{pmatrix}
      \phi_1\\
      \phi_2
    \end{pmatrix}$$
However:
$$ S_1=e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma}
 \begin{pmatrix}
      m_1 & 0 \\
      0 & m_2 
    \end{pmatrix} = \begin{pmatrix}
      m_1e^{-\frac{i\alpha_3}{2}} & m_2e^{-\frac{i(\alpha_1-i\alpha_2)}{2}} \\
      m_1 e^{-\frac{i(\alpha_1+i\alpha_2)}{2}}& m_2e^{\frac{i\alpha_3}{2}} 
    \end{pmatrix}$$
$$
 S_2=\begin{pmatrix}
      m_1 & 0 \\
      0 & m_2 
    \end{pmatrix}  e^{-\frac{i}{2}\vec\alpha \cdot \vec\sigma} =\begin{pmatrix}
      m_1e^{-\frac{i\alpha_3}{2}} & m_1 e^{-\frac{i(\alpha_1-i\alpha_2)}{2}} \\
      m_2 e^{-\frac{i(\alpha_1+i\alpha_2)}{2}} & m_2e^{\frac{i\alpha_3}{2}} 
    \end{pmatrix}$$
So $S_2=S_1^{T}$.
 A: The $\text{SU}(2)$ symmetry of this theory is only preserved when $m_1=m_2\equiv m$, in which case $M=m\textbf{1}$. Otherwise, the internal $\text{SU}(2)$ symmetry is broken down to $\text{U}(1)\times\text{U}(1)$, one for each scalar field. Something similar to this happens in QCD, in which the up and down quarks have nearly the same mass, and the approximate $\text{SU}(2)$ isospin symmetry is useful for classifying low-mass mesons and baryons.
A: Ignore all normalizations, since this is an issue of symmetry, and consider real numbers $a,b,c,d$, s.t.
$$
\phi_1=a+ib, \qquad  \phi_2=c+id ,
$$
and define $m_1=m_2 + \Delta$. 
Then, for the real 4-vector $\vec \varphi\equiv (a,b,c,d)^T$, you have
$$\mathcal{L}=\partial_{\mu} \vec \varphi \cdot \partial^{\mu} \vec \varphi -m_2  \vec \varphi \cdot  \vec \varphi
-\Delta (a^2+b^2).
$$
It is manifest that, for $\Delta=0$, this expression is SO(4) ~ SU(2)×SU(2) invariant. (One of these two SU(2) s is the SU(2) you display, but the other one is less easy to see in your language, and corresponds to the "right custodial" SU(2) of the SM, a global approximate symmetry of it. Through SO(4), you can appreciate there are 6 transformations mixing your 4 real scalars.)
But the introduction of the explicit perturbation   $\Delta$ breaks this SO(4) into $U(1)\times U(1) \sim O(2) \times O(2)$, the two O(2) s rotating the doublets (a,b) and (c,d), independently, respectively.  The other 4 generators of SO(4) are explicitly broken, since you cannot preserve the last, perturbation term, any other way--try it. 
