Create equal aspect (square) plot with multiple axes when data limits are different?

One way to deal with the problem is to keep the data limits of the x and y axis equal. This can be done by normalising the values to be between, say, 0 and 1. This way the command ax.set_aspect('equal') works as expected. Of course, if one only does this, the tick labels will only range from 0 to 1, so one has to apply a little matplotlib magic to adjust the tick labels to the original data range. The answer here shows how this can be accomplished using a FuncFormatter. However, as the original ticks are chosen with respect to the interval [0,1], using a FuncFormatter alone will result in odd ticks e.g. if the factor is 635 an original tick of 0.2 would become 127. To get 'nice' ticks, one can additionally use a AutoLocator, which can compute ticks for the original data range with the tick_values() function. These ticks can then again be scaled to the interval [0,1] and then FuncFormatter can compute the tick labels. It is a bit involved, but in the end it only requires about 10 lines of extra code:

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.ticker as mticker

from mpl_toolkits.axes_grid1 import make_axes_locatable

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)


fig,ax=plt.subplots()

divider = make_axes_locatable(ax)


##increased pad from 0.1 to 0.2 so that tick labels don't overlap
xhax = divider.append_axes("top", size=1, pad=0.2, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.2, sharey=ax)

##'normalizing' x and y values to be between 0 and 1:
xn = (x-min(x))/(max(x)-min(x))
yn = (y-min(y))/(max(y)-min(y))

##producinc the plots
ax.scatter(xn, yn)
xhax.hist(xn)
yhax.hist(yn, orientation="horizontal")

##turning off duplicate ticks (if needed):
plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

ax.set_aspect('equal')


##setting up ticks and labels to simulate real data:
locator = mticker.AutoLocator()

xticks = (locator.tick_values(min(x),max(x))-min(x))/(max(x)-min(x))
ax.set_xticks(xticks)
ax.xaxis.set_major_formatter(mticker.FuncFormatter(
    lambda t, pos: '{0:g}'.format(t*(max(x)-min(x))+min(x))
))

yticks = (locator.tick_values(min(y),max(y))-min(y))/(max(y)-min(y))
ax.set_yticks(yticks)
ax.yaxis.set_major_formatter(mticker.FuncFormatter(
    lambda t, pos: '{0:g}'.format(t*(max(y)-min(y))+min(y))
))

fig.tight_layout()
plt.show()

The resulting picture looks as expected and stays square-shaped also upon resizing of the image.

Old Answer:

This is more a workaround than a solution:

Instead of using ax.set_aspect(), you can set up your figure such that it is a square by providing figsize=(n,n) to plt.subplots, where n would be the width and height in inches. As the height of xhax and the width of yhax are both 1 inch, this means that ax becomes square as well.

import numpy as np
import matplotlib.pyplot as plt

from mpl_toolkits.axes_grid1 import make_axes_locatable

x = np.random.normal(512, 112, 240)
y = np.random.normal(0.5, 0.1, 240)

fig, ax = plt.subplots(figsize=(5,5))

divider = make_axes_locatable(ax)

xhax = divider.append_axes("top", size=1, pad=0.1, sharex=ax)
yhax = divider.append_axes("right", size=1, pad=0.1, sharey=ax)

ax.scatter(x, y)
xhax.hist(x)
yhax.hist(y, orientation="horizontal")

##turning off duplicate ticks:
plt.setp(xhax.get_xticklabels(), visible=False)
plt.setp(yhax.get_yticklabels(), visible=False)

plt.show()

The result looks like this:

Of course, as soon as you resize your figure, the square aspect will be gone. But if you already know the final size of your figure and just want to save it for further use, this should be a good enough quick fix.