Plane partitions with equal margins A plane partition of $n$ is an table of integers $A=(a_{ij})$ which add up to $n$ and non-increase in rows and columns. For example,
$$A= \begin{matrix} 331 \\
32 \ \  \\
11 \ \ 
\end{matrix}
$$
is a plane partition of $(3+3+1)+(3+2)+(1+1)=14$. 
One can view plane partitions as an arrangement of cubes stacked in the corner, see more on Wikipedia. 
Define margins to be 1-dim projections of cubes on all coordinate axis. These margins are triples $(\lambda,\mu,\nu)$ of partitions of $n$. For example, for $A$ as above we have $\lambda = (7,5,2)$, $\mu=(7,6,1)$ and $\nu=(7,4,3)$. 
Question: What is the smallest $n$ for which there exist two different plane partitions of $n$ with the same margins $(\lambda,\mu,\nu)$?  
Note: I know there are two different plane partitions of $2100$ with equal margins.  This is a consequence of $p_2(n)<p(n)^3$ for $n\ge 2100$, where $p_2(n)$ is the number of plane partitions.  Unfortunately, 2100 is way too large, and $p_2(2100)\approx 1.47\cdot 10^{141}$.  So it would be nice to find a small explicit example or an argument why e.g. there is none, say, for $n\le 200$.
UPDATE (Jan 29): Thank you, John Machacek, Gjergji Zaimi and Brian Hopkins, and apologies for being unclear.  No, I don't worry for the symmetries, so John's first answer is the correct one.  It's so nice and clean and Gjergji's argument is so simple and convincing, there is not much to add other than the motivation.
This is related to the study of Kronecker coefficients and a bound by Ernesto Vallejo (here) which should be better known: 
$$g(\lambda,\mu,\nu)\ge p(\lambda,\mu,\nu)$$
where $g(\lambda,\mu,\nu)$ is the Kronecker coefficient and $p(\lambda,\mu,\nu)$ is the number of (labeled) plane partitions with margins $\lambda,\mu,\nu$.  So I became interested if this bound is ever effective.  From asymptotic considerations I can see that $g(\lambda,\mu,\nu)\ge p_2(n)/p(n)^3 = \exp\Theta(n^{2/3})$, but what are those partitions?  I then found an easy construction of such $\lambda,\mu,\nu$ and this many plane partitions as long as there is one example with $p(\lambda,\mu,\nu)\ge 2$.  So now my pathetic 2100 can be replaces with 13.  Nice.  I will link my paper here discussing these and other bounds once we put it on the arXiv (joint with Greta Panova). 
 A: I don't know if this is optimal, but here is a pair of plane partitions of $n=53$ with no symmetries that have equal projections on each individual axis:
$$A=\begin{matrix} 5 4 4 4 4 \\ 5 4 1 1 \ \ \\ 5 2 1 1 \ \ \\ 5 1 1 1 \ \ \\ 1 1 1 1\ \ \end{matrix}$$
and
$$B=\begin{matrix} 5 5 5 5 1 \\ 4 4 1 1 1  \\ 4 2 1 1 1  \\ 4 1 1 1 1  \\ 4 \ \ \ \ \ \ \  \ \end{matrix}$$
This is, of course, a bit like cheating because we just overlapped two cyclically symmetric partitions that are transpose with an L shape to break any symmetries.

If the question doesn't care about symmetries, and is simply asking for the smallest $n$ for which some plane partition of $n$ is not uniquely determined by the projection triple $(\lambda,\mu,\nu)$ then John Machacek's example is the minimal one. 
Proposition: For $n\le 12$ a plane partition of $n$ is uniquely determined by the triple $(\lambda, \mu, \nu)$.
Proof: It is simple to check that for a classical partition $\alpha$, if $\alpha$ is not determined by the length of its longest row and longest column, $(\alpha_1, \alpha'_1)$, then we must have $\min(\alpha_1,\alpha'_1)\geq 3$ and $|\alpha|-\alpha_1-\alpha'_1\geq 1$. 
Now let's denote the rows of our plane partition as $\beta^1, \dots, \beta^r$ where each $\beta^i$ is a partition. If $\beta^1$ doesn't satisfy the conditions of the previous paragraph then we could reconstruct $\beta^1$ from the knowledge of its longest row and column. Both of these can be found from the information of the projections of the plane partition. Once we reconstruct $\beta^1$ we can repeat the process and reduce the problem to reconstructing $\beta^2, \dots, \beta^r$. Eventually we recover the whole partition. Moreover we can read the plane partition using a different coordinate plane and obtain the same information for $\gamma^1$ and $\delta^1$ which denote the slices of our partition on the planes $y=0$ and $z=0$.
Since each of $\beta^1, \gamma^1, \delta^1$ satisfy the conditions of the first paragraph we have that the size of our plane partition is at least 
$$|\beta^1|+|\gamma^1|+|\delta^1|-\frac{1}{2}((\beta^1)_1+(\beta^1)'_1+(\gamma^1)_1+(\gamma^1)'_1+(\delta^1)_1+(\delta^1)'_1)+1$$
$$\geq |\beta^1|+|\gamma^1|+|\delta^1|-((\beta^1)_1+(\beta^1)'_1+(\gamma^1)_1+(\gamma^1)'_1+(\delta^1)_1+(\delta^1)'_1)+10\geq 13$$
as desired.
A: Hopefully I understood the definitions and computed correctly. How about $n = 13$ with
$$A = \begin{matrix} 3 3 1 \\ 2 1 1 \\ 2 \ \ \end{matrix}$$
and
$$B = \begin{matrix} 3 2 2 \\ 3 1 \ \\ 1 1 \ \end{matrix}$$
where all margins are $(7,4,2)$.
A: Here is a smaller example than Gjergji Zaimi's that has at least less symmetry than John Machacek's.  Among the plane partitions of 14,
$$ A = \; \begin{matrix} 3 & 2 & 2 & 1 \\ 3 & 1 \\ 1 & 1 \end{matrix} $$
and
$$ B = \; \begin{matrix} 3 & 3 & 1 & 1 \\ 2 & 1 & 1 \\ 2 \end{matrix} $$
both have margins $((8,4,2), (7,4,2,1), (8,4,2))$.
The smallest pairs with equal distinct margins occur with $n = 16$, e.g., 
$$ C = \; \begin{matrix} 5 & 2 & 2 & 1 \\ 3 & 1 \\ 1 & 1 \end{matrix} $$
and
$$ D = \; \begin{matrix} 5 & 3 & 1 & 1 \\ 2 & 1 & 1 \\ 2 \end{matrix} $$
both have margins $((10, 4, 2), (9, 4, 2, 1), (8, 4, 2, 1, 1))$.
(I was able to get Mathematica to check for equal margins among all plane partitions through $n = 16$ and might be able to go a little farther.  Luckily these examples came up within the bounds of my programming ability and computer power.)
