The $\sigma_i$ as defined above satisfy $\mathrm{d}\sigma_i = -2\,\sigma_j\wedge\sigma_k$ when $(i,j,k)$ is an even permutation of $(1,2,3)$. For later use, let $E_i$ be the dual (left-invariant) frame field on $\mathrm{SU}(2)$.
Using these structure equations, one can compute that, if
$$
g = A_1\,{\sigma_1}^2+A_2\,{\sigma_2}^2+A_3\,{\sigma_3}^2
$$
where the $A_i$ are positive constants, then
$$
\mathrm{Ric}(g) = B_1A_1\,{\sigma_1}^2+B_2A_2\,{\sigma_2}^2+B_3A_3\,{\sigma_3}^2,
$$
where, for $(i,j,k)$ any permutation of $(1,2,3)$,
$$
B_i = \frac{2(A_i+A_j-A_k)(A_i-A_j+A_k)}{A_1A_2A_3}.
$$
The $B_i$ are the eigenvalues of the Ricci tensor of $g$.
If $B_1$, $B_2$, and $B_3$ are distinct, then, since any isometry of $g$ must preserve $\mathrm{Ric}(g)$, it must preserve each of the $\sigma_i$ up to sign. By the structure equations, any map $\iota:{\mathbb{S}}^3\to {\mathbb{S}}^3$ that preserves each $\sigma_i$ up to sign must satisfy $\iota^*(\sigma_i) = \epsilon_i\sigma_i$ where $\epsilon_i^2=1$ and $\epsilon_1\epsilon_2\epsilon_3 = 1$. The group $G_0$ of such maps has four components, with the identity component being the left-translations. The other three components are given by left translations combined with right multiplication by the $X_i\in\mathrm{SU}(2)$. The group $G_0$ is a subgroup of the isometry group of $g$ for any choice of the $A_i>0$.
Since $B_i-B_j = -4(A_i{-}A_j)(A_k{-}A_i{-}A_j)/(A_1A_2A_3)$, if $B_i=B_j$, then either $A_i=A_j$ or $A_i+A_j=A_k$.
If $B_1=B_2=B_3$, then $A_1=A_2=A_3$, and the metric $g$ has constant sectional curvature. Up to a scalar multiple, it is the standard round sphere, so the group of isometries of $g$ is $\mathrm{O}(4)$.
If only two of the $B_i$ are equal, then for some $(i,j,k)$ distinct, one has $B_i=B_j\not=B_k$, so either $A_i=A_j\not=A_k$ or $A_i \not= A_j$ while $A_k = A_i + A_j$.
When $A_i=A_j\not=A_k$, the isometry group of $g$ has to preserve ${\sigma_i}^2+{\sigma_j}^2$ and ${\sigma_k}^2$. From this, it follows that the isometry group $G_k$ of $g$ has dimension $4$ and is generated by $G_0$ plus the flow of the left-invariant vector field $E_k$ that is dual to $\sigma_k$. The group $G_k$ has two components, since the flow of $E_k$ connects the two components of $G_0$ whose elements leave $\sigma_k$ fixed.
When $A_i\not= A_j$ and $A_i+A_j = A_k$, the argument is a little more complicated, but the isometry group in this case is just the $3$-dimensional group $G_0$ (with four components) that preserves each of the $\sigma_i$ separately up to sign. The point is that the isometry group of $g$ has to preserve both $A_i\,{\sigma_i}^2 + A_j\,{\sigma_j}^2$ and ${\sigma_k}^2$. In particular, it has to preserve the vector field $E_k$ up to sign and hence preserve up to sign the Lie derivative of $A_i\,{\sigma_i}^2 + A_j\,{\sigma_j}^2$ with respect to $E_k$, which is $2(A_j{-}A_i)\sigma_i\sigma_j$. Thus, if an element $\iota$ of the isometry group of $g$ did not belong to $G_0$, it would have to preserve ${\sigma_k}^2$ and satisfy $\iota^*({\sigma_i}) = \lambda\,{\sigma_j}$ and $\iota^*({\sigma_j}) = \pm\lambda^{-1}\,{\sigma_i}$ for some nonvanishing function $\lambda$. This would imply
$\iota^*(\sigma_i\wedge\mathrm{d}\sigma_i)
=\iota^*\sigma_i\wedge \mathrm{d}(\iota^*\sigma_i) = \lambda^2\,\sigma_j\wedge\mathrm{d}\sigma_j\,.
$
Since $\sigma_i\wedge\mathrm{d}\sigma_i
=\sigma_j\wedge\mathrm{d}\sigma_j = -2\,\sigma_1\wedge\sigma_2\wedge\sigma_3$ is a constant multiple of the volume form of $g$, which $\iota^*$ must preserve up to sign, it follows that $\lambda^2=1$. Thus, such an $\iota$ would pullback $A_i\,{\sigma_i}^2 + A_j\,{\sigma_j}^2$ to $A_i\,{\sigma_j}^2 + A_j\,{\sigma_i}^2$ and hence would not be an isometry of $g$.
