If we take units in which the speed of light is 1, the Lorentz transformation is exactly symmetric in $#x#$ and $#t#$.
So why do we perceive the results as length contraction and time dilation? (ie as seeing the other frame with shorter measuring rods and longer clock ticks)
To see the answer here we need to understand the asymmetry between how we compare measuring rods and clock tick intervals.
When we observe the length of another observer’s measuring rod we do so by looking at the spatial coordinates of both its endpoints at the same time in our frame. If the ends are 1 unit apart in the traveller’s frame then we have $#\Delta x’=1#$ and so, since $#\Delta x’=\gamma_{v}(\Delta x + v\Delta t)#$, when $#\Delta t=0#$ we have $#\Delta x=1/\gamma_{v}#$. ie the length as seen by us is less and we see the traveller’s measuring rod as “contracted”.
The symmetric case for a time interval would be to measure both ends at the same position in our frame.
But when we count clock ticks they are not at the same position in our frame but rather in the travelling frame. The interval we observe corresponding to one unit of time for the traveller, ie for $#\Delta t’=1#$, is the time in our frame between the events in the traveller’s frame with $#\Delta x’=0#$ (rather than with $#\Delta x=0#$ which would be the analogue of what we did with the measuring rods).
This gives $#0=\gamma_{v}(\Delta x + v\Delta t)#$ so $#\Delta x=-v\Delta t#$, and so $$\Delta t’=\gamma_{v}(\Delta t – (v/c^2)\Delta x)=\gamma_{v}(1-(v^2/c^2))\Delta t=(1/\gamma_{v})\Delta t$$ So $#\Delta t=\gamma_{v}\Delta t’#$ which means that our clock ticks more frequently and we see the moving one as slowed down (and so having a longer “dilated” tick interval).
So what would the symmetric situation correspond to?
Well if we looked at events in the travelling frame with $#\Delta x=0#$ they would not be at the same position in the traveller’s frame and it would be like the traveller having a string of coordinated flashing lights and us measuring the time interval between two flashes at our own location rather than two flashes of the same travelling light (and since simultaneity is different for the traveller, the clocks or lights that she had synchronized are not in fact synchronized for us).
And going back to lengths, if we determined the length of the measuring rod by looking at our spatial coordinates of both its endpoints at the same time in its own frame, then we’d actually be looking at the back end a bit earlier in our frame and so it’s front would have travelled on a bit and instead of being contracted we would indeed consider it dilated.
But to be frank, I’m still not satisfied with this answer. I’d like to see a more intuitive explanation of why we take this asymmetric approach to identifying the moving frame’s intervals of space (by fixing time in our frame) and time (by fixing position in the other frame).