In 2-d space, it is possible to take any shape and fit it inside a square such that it touches all the sides of the square. In other words, its projection on the x-axis is the same as its projection on the y-axis. To see this, we can consider a function which is the ratio of its projection on the x-axis to its projection on the y-axis. As we rotate by $\frac{\pi}{2}$, this ratio will go from one side of $1$ to the other. Because it is a continuous function, it must cross $1$.
I was wondering about extending this result to 3-d space. I have a general 3-d object and would like to rotate it such that its projection along all three axes becomes the same. Can we prove this is always possible or otherwise? And if it isn't possible in general, any non-trivial special cases where it will be?
Proof of the conjecture
Short sketch of proof. For every choice of an axis of rotation, one can spin the body around that chosen axis. As the OP noted, one can make the two perpendicular dimensions in this plane match. Unfortunately, the thickness of the body in the third direction, measured along the chosen axis, might be too large or too small. In fact we know the third dimension can be made too large/too small if we point the axis in the direction of (A) absolutely largest/ (B) absolute smallest possible thickness of the body. If we gradually move the axis of rotation from A to B there must be an intermediate Goldilocks choice $C$ that is just right.
Details of proof.
For every unit vector $\vec n$ on the unit sphere $S$, denote the thickness of the body in that direction by $h(\vec n)$. (The thickness is the length of the segment obtained by projecting the body orthogonally onto the line whose direction is $\vec n$.) This is a non-negative continuous function of $\vec n \in S$. Note that $h(-\vec n)=h(\vec n)$.
The continuous function $h(\vec n)$ attains its absolute max at least once at some direction; call one such choice $A\in S$. Likewise its absolute minimum is attained somewhere; call one such choice $B\in S$. Construct a continuous path on the sphere from $A$ to $B$. Let $C$ be any intermediate point on that path.
Regarding $C$ as the vertical axis pointing upward out of our field of view, the directions perpendicular to $C$ sweep out a circle of directions in the plane parametrized by an angle $\theta$. We want to show that there is a solution of $f_1(\theta)=h(\theta)= f_2(\theta)=h(\theta+\pi/2)$. (This provides a pair of perpendicular directions of equal width in the plane).
Technical point: in order to single out an attribute of the solutions that can be shown to be continuous as we vary $C$, some extra care is needed. The problem is that the domain point $\theta$ might not be unique or might not depend continuously on $C$. In contrast, the maximal range value we construct below will depend continuously on $C$.
Define $f_3(\theta)=min(f_1(\theta), f_2(\theta))$
The two planar regions defined in polar coordinates as $r\leq f_1(\theta)$ and $r\leq f_2(\theta)$ respectively are two congruent regions that are rotated by a right angle. Each of these congruent regions is defined as an intersection of half-planes, hence the region a convex region in the plane. The intersection of these two convex congruent regions is another convex region whose polar equation is $r= f_3(\theta)$. (see picture below). The largest value of $f_3(\theta)$ will be denoted by $r_{max}(C)$. It is the largest possible value of $r$ for which $r= h(\theta)= h(\theta+\pi/2)$. Continuity of $r_{max}(C)$ follows from the joint continuity of the expression $f_3$ as a function of $\theta$ and $C$ on the compact configuration space $(\theta, C)$.)
Note that by the choice of $A$ as the absolutely largest value of $h(\vec n)$ on the entire sphere, it follows that $r_{max}(A) \leq h(A)$.
At the end of the path, by construction of $B$ as a point where $h(\vec n)$ is absolutely smallest, we must have $r_{max}(B)\geq h(B)$. By the intermediate value theorem, there is a point $C$ where $r_{max}(C)= h(C)$. At this point $C$ we have the equality of the thickness function $h$ in three perpendicular directions. QED.
P.S. For a nontrivial class of extreme examples for which equal widths do occur in every direction, you can explore the topic of surfaces of constant width The obvious example is a sphere. Remarkably, however, there exist many non-spherical convex smooth surfaces with this property: the projected width of the object on each line of projection is constant in every direction chosen for the line.
This is an outline of how aspherical constant-width surfaces are constructed. Starting with a desired function $h(\vec n)$ one parametrizes the desired surface as $\vec p= h(\vec n) \vec n + \vec t(\vec n)$ where the correction term $\vec t$ is perpendicular to $\vec n$. It turns out that the correct choice for the correction is $\vec t=\nabla_S h$ where $\nabla_S h$ denote the spherical gradient of $h$, (the gradient of the function $h$ treated as a function on the unit sphere. To create a constant width surface that is a perturbation of the round sphere, of radius $R$, set $h= R+ \epsilon(\vec n)$ where the perturbation is a small odd function of $\vec n$.
Then the total width in direction $\vec n$ is $h(\vec n) + h(-\vec n)=2R$. The messy details are checking that the perturbation $\epsilon$ is small enough to preserve convexity of the surface. That boils down to looking at the second derivatives of $\epsilon $ on the sphere.