How will it appear on a picture? Contracted or stretched? The first sentence in the article claims: 
https://en.wikipedia.org/wiki/Length_contraction
Length contraction is the phenomenon of a decrease in length of an object as measured by an observer who is traveling at any non-zero velocity relative to the object.
As I understand, If I move relative to my friend and have a photo camera and I will make a picture of my friend, he will appear contracted. He will appear of the same height, but will appear “very slim”, is it correct?
But if my friend will take picture, he will see that I am contracted. Is it correct?
But, doesn't a film in a camera Lorentz contract? If the film contracts, my friend will appear stretched. I will appear stretched too. How come?
If my friend holds two Einstein - synchronized clocks in left and right hands - will they show the same time on the picture? 
 A: Surprisingly, Lorentz contraction is not perceived by an observer as a contraction, but as a rotation. This is because the light rays that arrive at the observer simultaneously have not left the object at the same time, whereas Lorentz contraction is associated to the distance of the ends of moving objects at the same time in the observers frame of reference. 
I quote from [Terrell,1959]

A sphere will photograph with precisely the same circular outline, whether stationary or in motion with respect to the camera. An object with less symmetry than a sphere, such as a meter stick, will appear, when in rapid motion with respect to an observer, to have undergone rotation, not contraction.

See also [Penrose,1959] and the Wikipedia article on Terrell rotation.
A: So the first problem here is that you're trying to say "I make a picture of my friend" and the article is saying "length of an object as measured." Photographs are not a great way to measure distances. Go ahead and take a picture with your camera and then look very carefully at the picture and try to ask, "how would I say how long this thing is?". You will probably need to make reference to the things you see in the photograph, 'hey, this thing I know is 7 feet tall' and that will be difficult.
The second problem here is that your camera collects light waves, which move at the speed of light. When you start moving relative to your friend at speeds approaching the speed of light, your naive expectation of what's going to happen with those light waves is not necessarily correct. So to do this calculation you really need to say: 


*

*I have this friend showing me a picture of dots at some $x_i$ coordinates, she is at a closest-approach distance $y=L$ from me: in her coordinates, with me passing through the origin, their world-lines are "$(w,x,y,z) = (c\tau, x_i, L, 0),$ for all $\tau.$ 

*Now I do a Lorentz transformation to these lines in the $x$-direction and find the lines $(\gamma~(c\tau - \beta x_i), \gamma~(x_i - \beta c \tau), L,0) $ for all $\tau$ for each $i$. Furthermore if each point has a clock at it, $\tau$ is the time that appears on that clock.

*Now I know that in my reference frame I am at the origin and some light emitted by points on these two lines travels forward in time until it hits the aperture of my camera at some time $t$. If I say that the distance it travels divided by the time it travels must be $c$, via the Pythagorean theorem that amounts to: $$[c t - \gamma~(c\tau - \beta x_i)]^2 = L^2 + [\gamma~(x_i - \beta c \tau)]^2,$$ letting me find $\tau$ as a function of $t$ and $x_i.$

*Now that I know $\tau(t, x_i)$, I have to evaluate what the angles are between the 2D vectors $[L, \gamma (x_i - \beta c ~ \tau(t, x_i))]$ at constant $t$. Those angles correspond to the distances between these points that I'll see in the photograph. Alternatively I can try to project these light rays through an "image plane," probably one that is perpendicular to the light ray corresponding to $x_i = 0.$


Only after doing all of these steps can you be fully sure what the people in the spaceships see.
