This is actually a really interesting question and you will probably be surprised by some of the findings: for example, you might be surprised to know that as you accelerate faster and faster, the stars all "crowd into" the part of the sky in front of you. Even more spectacular, if I throw a beach ball past you, at no point does it present anything other than a circular profile to you: its length contraction is always "hidden" by other effects! These are called Terrell effects or "Terrell rotations" in general; the world as-you-see-it looks surprisingly sensible even though if you literally tried to calculate where everything was in your coordinate system you would have to ultimately conclude that things were squished in weird ways.
As for proving it, I only know of one good way, and that is the spinor calculus.
Adjoin the identity matrix $I$ to the usual Pauli matrices as $\sigma_w$ (where $w=ct$ as usual) and take the 4-vector $v^\mu$ of your choice in the $(w,~x,~y,~z)$ coordinates of your choice; now form the Hermitian matrix $V = v^\mu \sigma_\mu:$ now it turns out that $\det V = v^\mu v_\mu$ in the $(+~-~-~-)$ metric.
This has two consequences. First off null 4-vectors have $\det V = 0$ and therefore they are projections; $V = \phi~\phi^\dagger$ for some "spinor" $\phi$ living in $\mathbb C^2$. This is great because null vectors are precisely the light rays that are coming into your camera! To get a better sense of where they are, chart them starting from whenever they pass through a sphere of radius $R$ around you, then their 4-vector components are $(v^w,~v^x,~v^y,~v^z) = (-R, x, y, z),$ and the above procedure assembled them into some $\phi = [\alpha;\;\beta]$ according to: $$
V = \begin{bmatrix}v^w + v^z & v^x - i v^y\\
v^x + i v^y & v^w - v^z\end{bmatrix} = \begin{bmatrix}-R + z & x - i y\\
x + i y & -R - z\end{bmatrix} = \begin{bmatrix}\alpha~\alpha^* & \alpha~\beta^*\\
\beta~\alpha^* & \beta~\beta^*\end{bmatrix}.$$ Note that therefore $\zeta = -\beta/\alpha = -\beta\alpha^*/(\alpha\alpha^*) = -(x + i y)/(-R + z) = (x + i y)/(R - z),$ which is the stereographic projection of this point $(x, y, z)$ on this sphere of radius $R$ onto the complex plane $\mathbb C.$
So it's actually just fallen out of our maths that we want to describe these images by stereographically projecting whatever we see onto the complex plane!
I said above that there was also a second consequence; it is that the Lorentz transforms are linear transforms of $V$ which preserve $\det V$ and Hermitian-ness: aside from the parity transforms $V \to -V$ and $V \to (\det V)~V^{-1},$ you have the special Lorentz transforms $V \to L~ V~ L^\dagger$ for some complex matrix $L$ such that $\det L = 1.$
One such matrix is $$ L = \begin{bmatrix} e^{r/2} & 0 \\
0 & e^{-r/2} \end{bmatrix}.$$ This has the effect on our 4-vectors of $$ V \to \begin{bmatrix} e^{r/2} & 0 \\
0 & e^{-r/2} \end{bmatrix} \begin{bmatrix}v^w + v^z & v^x - i v^y\\
v^x + i v^y & v^w - v^z\end{bmatrix} \begin{bmatrix} e^{r/2} & 0 \\
0 & e^{-r/2} \end{bmatrix} = \begin{bmatrix}e^r(v^w + v^z) & v^x - i v^y\\
v^x + i v^y & e^{-r}(v^w - v^z)\end{bmatrix}, $$ but since $v^w = (V_{00} + V_{11})/2$ and $v^z = (V_{00} - V_{11})/2$ you can see that this is straightforwardly the Lorentz boost:$$\begin{array}{rl}
v^w\to&v^w \cosh r + v^z \sinh r\\
v^z\to&v^w \sinh r + v^z \cosh r, \end{array}$$making this the Lorentz boost in the $z$ direction by rapidity $-r$ or velocity $v = -c \tanh r.$ In other words this is the transform you use when you transfer to a reference frame moving relative to you in the $-z$ direction.
Similarly we find that $L \phi = [e^{r/2}~\alpha;\; e^{-r/2} \beta]$ and so this maps the images that we're seeing on the complex plane by $\zeta \to e^{-r} \zeta,$ so they all crowd together towards the origin point $\zeta = 0.$ But looking at "what is $\zeta = 0$ on the original sphere?" we have that $x, y \to 0$ while $R - z \to \text{nonzero},$ hence it is the point $(0, 0, -R),$ proving that indeed when you change these coordinates the stars all crowd in the direction that you're going.
Similarly you can use this to calculate the angle-distortion in your photographs; it's as simple as "stereographically project around the direction you're going; scale $\zeta \to k \zeta$ for some constant $k < 1$, then stereographically project back."
(It is also instructive to consider future-pointing null vectors, in which case the formula for the stereographic projection is different and you derive a similar effect; the light that you are emitting in all directions is concentrated in the direction you're seen as going from some third-party perspective. This is somewhat more well-known than the Terrell effects and is called relativistic beaming.)
As for that ball? Well, stereographic projections map circles on the sphere to circles and straight lines (for circles passing through $(0, 0, R)$) and vice versa. The action of any such matrix $L$ upon $\zeta$ is a bilinear or "Mobius" transform which also maps circles and lines to circles and lines. So therefore every Lorentz transform preserves every circle that you visually see, even though they may distort a pattern of uniformly nested circles (producing the Terrell rotation effect and other apparent skews).
