Summary: One possible transformation is $T(z) = 2\frac{z+3}{z-3}$, which maps the region in question onto the region $\{w\in\mathbb{C}:1<|w|<4\}$.
Let $T$ be a desired transformation, and let $a = T^{-1}(0)$ and $b = T^{-1}(\infty)$. Then $T(z) = c\frac{z-a}{z-b}$ for some $c\in\mathbb{C}$. From symmetry considerations, we can take $a$ and $b$ to be real. For $T$ to map the circle $C_1 = \{z\in\mathbb{C}:|z+5|\le 4\}$ to a circle on the origin, we must have that $|T(z)|$ is constant on $C_1$. In particular, since $-1\in C_1$ and $-9\in C_1$, it follows that $|T(-1)| = |T(-9)|$. Thus,
$$\left|c\frac{-1-a}{-1-b}\right| = \left|c\frac{-9-a}{-9-b}\right|\implies \left|\frac{1+a}{1+b}\right| = \left|\frac{9+a}{9+b}\right|.$$
Since $T$ is injective, it follows that
$$ \frac{1+a}{1+b} = \frac{T(-1)}{c}\ne\frac{T(-9)}{c} = \frac{9+a}{9+b}.$$
Hence, $\frac{1+a}{1+b}$ and $\frac{9+a}{9+b}$ are distinct real numbers with the same magnitude, so they must be additive inverses of each other. Hence, $\frac{1+a}{1+b} = -\frac{9+a}{9+b}$, and after a few manipulations we obtain the equation $18 + 10a + 10b + 2ab = 0$.
Applying the same logic to the circle $C_2 = \{z\in\mathbb{C}:|z-5|\le 4\}$ yields the relation
$$ \frac{1-a}{1-b} = -\frac{9-a}{9-b}\implies 18 - 10a - 10 b + 2ab = 0.$$
Subtracting the two equations yields $a+b=0$, i.e. $b=-a$. Substituting this into the first equation yields $18 - 2a^2 = 0\implies a^2 = 9\implies a = \pm 3$. Hence, we can take $a=-3$ and $b=3$, yielding $T(z) = c\frac{z+3}{z-3}$. In this case, $C_1$ should be mapped onto a circle of radius $1$ around the origin, and $C_2$ should be mapped onto a circle of radius $R>1$ around the origin, since $T^{-1}(0)$ is in the interior of $C_1$ and $T^{-1}(\infty)$ is in the interior of $C_2$.
For $T$ to map $C_1$ to a circle of radius $1$ around the origin, we need $|T(z)| = 1$ for all $z\in C_1$. In particular, $|T(-1)| = 1$, i.e.
$$\left|c\frac{-1+3}{-1-3}\right| = 1\implies \left|\frac{c}{2}\right| = 1.$$
Hence, we can take $c = 2$. Now, to show that $T$ does map $C_1$ to a circle of radius $1$ around the origin, it suffices to find three points $z_1,z_2,z_3\in C_1$ such that $|T(z_i)| = 1$, since we know that $T(C_1)$ is a circle, and a circle is uniquely determined by any three of its points. From the above calculations, we already know that $|T(-1)| = |T(-9)| = 1$. We can choose the third point to be $z = -5+4i$, and we can check:
$$|T(-5+4i)| = \left|2\frac{(-5+4i)+3}{(-5+4i)-3}\right| = 2\left|\frac{-2+4i}{-8+4i}\right| = 2\frac{\sqrt{2^2+4^2}}{\sqrt{8^2+4^2}} = 1.$$
Thus $T$ maps $C_1$ to $\{w\in\mathbb{C}:|w| = 1\}$. Similar calculations show that $T$ maps $C_2$ to a circle around the origin. To calculate the radius of that circle, it suffices to evaluate $T$ somewhere on $C_2$, e.g. at $z=1$, yielding
$$ R = |T(1)| = \left|2\frac{1+3}{1-3}\right| = 4. $$
Thus, $T(z) = 2\frac{z+3}{z-3}$ maps the circles $C_1$ and $C_2$ to the circles around the origin of radii $1$ and $4$, so it maps the region between the circles (i.e. the region in question) to the annulus around the origin of radii $1$ and $4$.
