We will solve this problem with the Polya Enumeration Theorem. Evidently this is equivalent to counting two-colorings of the small cubes (leaving out a cube is just like coloring it a second, different color). This requires the cycle index of the permutation group $G$ of the small cubes inside the large one. It is important to note that we restrict ourselves to rotations. If we had a graph instead of a real-world object there would be mirror reflections to consider as well (reflections about a plane passing through the center of the cube and parallel to two opposite sides). This cannot be done with a physical concrete object. We now enumerate the permutations of $Z(G)$ by their cycle structure, considering the types of rotations in turn.
Start with the identity, which contributes
$$a_1^{27}.$$
Now there are eight rotations about axes passing through opposite corners of the cube (two rotations per axis, of which there are four), which fix the three cubes on the diagonal and which contribute
$$8\times a_1^3 a_3^8.$$
There are three rotations about axes passing through the centers of opposite faces, which again fix the cubes on said axis and which give $$3\times (2 a_1^3 a_4^6 + a_1^3 a_2^{12}).$$
Finally we have six rotations about axes passing through the centers of opposite edges, which fix the cubes on that axis and give
$$6\times a_1^3 a_2^{12}.$$
This gives the cycle index
$$Z(G) = \frac{1}{24}
\left(a_1^{27} + 8\times a_1^3 a_3^8 + 3\times (2 a_1^3 a_4^6 + a_1^3 a_2^{12})
+ 6\times a_1^3 a_2^{12}\right)$$
which is
$$Z(G) = \frac{1}{24}
\left(a_1^{27} + 8\times a_1^3 a_3^8 + 6\times a_1^3 a_4^6
+ 9\times a_1^3 a_2^{12}\right).$$
Now this gives
$$Z(G)(1+z) =
1/24\, \left( 1+z \right) ^{27}+1/3\, \left( 1+z \right) ^{3} \left( 1+{z}^{3}
\right) ^{8}\\+1/4\, \left( 1+z \right) ^{3} \left( 1+{z}^{4} \right) ^{6}+3/8\,
\left( 1+z \right) ^{3} \left( 1+{z}^{2} \right) ^{12}$$
which is
$$Z(G)(1+z) =
{z}^{27}+4\,{z}^{26}+22\,{z}^{25}+139\,{z}^{24}+779\,{z}^{23}+3455\,{z}^{22}\\+12507\,
{z}^{21}+37303\,{z}^{20}+92968\,{z}^{19}+195963\,{z}^{18}+352433\,{z}^{17}+544382\,{
z}^{16}\\+725612\,{z}^{15}+837184\,{z}^{14}+837184\,{z}^{13}+725612\,{z}^{12}+544382\,
{z}^{11}+352433\,{z}^{10}\\+195963\,{z}^{9}+92968\,{z}^{8}+37303\,{z}^{7}+12507\,{z}^{
6}+3455\,{z}^{5}\\+779\,{z}^{4}+139\,{z}^{3}+22\,{z}^{2}+4\,z+1,$$
so the answers are four, twenty-two, and one hundred thirty nine respectively.
This MSE link computes the full face permutation group of the n-dimensional hypercube.
I will show how to solve this problem when the base is an $n$-dimensional cube and the apex is directly above a point in the base. Then I will show how the procedure can be generalized if the cube is replaced with an arbitrary convex polyhedral figure. I would suggest you draw pictures for the arguments below in the case $n = 2$.
Say the $n$-dimensional cube is defined by $2n$ inequalities $x_i \geq - 1, x_i \leq 1$, for $i = 1, \dots, n$, and the apex is located at $c = (c_1,\dots,c_n,c_{n+1})$, where $\hat{c} = (c_1,\dots,c_n,0)$ is a point in the interior of the base. That means we're assuming the $c_i$'s satisfy the strict inequalities corresponding to the lax ones above, i.e., $c_i > -1, c_i < 1$ for all $i = 1, \dots, n$.
Now we let $\hat{x} = (x_1,\dots,x_n,0)$ be any point in the base and our problem is to find the value of $x_{n+1}$ such that the point $x = (x_1,\dots,x_n,x_{n+1})$ will belong to the pyramid.
The pyramid has one $n$-dimensional face for each $(n-1)$-dimensional face of the base cube. Our first task is to determine which face corresponds to the face $x$ is located on. To do this, we draw the ray originating at $\hat{c}$ and passing through $\hat{x}$. That ray hits one of the faces. That face is the one we want, or at least the point on it where the ray meets it. The ray consists of all points $\hat{c} + t(\hat{x} - \hat{c})$ as $t$ ranges through values $t \geq 0$ (assuming $\hat{x} \ne \hat{c}$). The ray hits a face when $t$ is the smallest possible positive value for which one of the inequalities defining the cube becomes an equality. In other words, we look at the $2n$ equations $c_i + t(x_i - c_i) = \pm 1$, we solve them all for $t$, and whichever one has the smallest positive solution for $t$, which we'll name $T$, is the face we want. (We must in fact have $T \geq 1$.) Now we write $y$ for the point where the ray first meets a face. We have $y = (c_1 + T(x_1 - c_1),\dots,c_n + T(x_n - c_n),0)$.
Now it's easy to find $x$. If we draw the triangle whose vertices are $c$, $\hat{c}$ and $y$, we see that it's a right triangle with a right angle at $\hat{c}$, with the side from $\hat{c}$ to $c$ vertical and from $\hat{c}$ to $y$ horizontal. Then $\hat{x}$ is a point on the horizontal leg of the triangle, $1/T$ of the way from $\hat{c}$ to $y$, and $x$ lies directly above it on the hypotenuse. We therefore have $x = c + (1/T)(y - c)$, so $x_{n+1} = (1 - 1/T)c_{n+1}$.
In summary, solve all the equations $c_i + t(x_i - c_i) = \pm 1$ for $i = 1,\dots, n$, and let $t = T$ be the smallest solution that is positive. Then we have $x_{n+1} = (1 - 1/T)c_{n+1}$. An exceptional case occurs when $\hat{x} = \hat{c}$, and then $x_{n+1} = c_{n+1}$.
More generally, we can repeat the above arguments and procedure almost exactly when we have a convex base defined by several linear inequalities of the form $a_1 x_1 + \dots a_n x_n \leq b$, and the apex is directly above a point in the interior of the base. Geometrically, the convexity condition means that if you take any $(n-1)$-dimensional face of the base, then the entire base lies on the same side of that face. There are only two major differences with the special case above.
The first difference is that some of the inequalities may be superfluous. That is, if we remove those conditions, we still obtain the same set. In that case the "face" corresponding to that particular inequality will have dimension lower than $n$ and will be contained in one or more of the real $n$-dimensional faces. This makes no difference in how you calculate $x_{n+1}$.
The other difference is that it may happen that you don't have enough linear inequalities to make the set defined by them bounded. In other words, the set might not be contained within a region of space that has finite dimensions. In this situation, I don't think the concept of a pyramid would make sense. At the very least, it would be very different, and the above procedure wouldn't work.
Edit: Here are the formulas for the general case. I'll simplify the formula a bit by going straight to $1-1/T$ rather than going through the step of calculating $T$. Assume your $n$-dimensional base is defined by several linear inequalities $a_1 x_1 + \dots a_n x_n \leq b$, and that the coordinates of the apex $(c_1,\dots,c_n,c_{n+1})$ satisfy each inequality $a_1 c_1 + \dots a_n c_n < b$. Now let a point $\hat{x} = (x_1,\dots,x_n,0)$ be given in the base (i.e, all the lax inequalities are satisfied by its coordinates). Now, for each inequality defining the base, calculate the quantity
$$s = 1 - 1/t = \frac{b-a_1 x_1 - \dots - a_n x_n}{b - a_1 c_1 - \dots - a_n c_n},$$
and let the smallest of these numbers (which are all nonnegative) be $S = 1 - 1/T$. Then $x_{n+1} = Sc_{n+1}$. ($\hat{x} = \hat{c}$ is no longer an exceptional case.)
In the special case of a cube, the $2n$ values are $s = \frac{1 + x_i}{1 + c_i}$ and $s = \frac{1 - x_i}{1 - c_i}$ for $i = 1,\dots,n$, and $S$ is the minimum of these.
Edit: Here is an illustration of the right triangle mentioned in the solution. Ray $a$ starts at $\hat{c}$ when $t = 0$, passes through $\hat{x}$ when $t = 1$, and meets the boundary of the base at $y$ when $t = T$. Hence the distance from $\hat{c}$ to $\hat{x}$ is $1/T$ times the distance from $\hat{c}$ to $y$. The points $c, \hat{c}, y$ form a right triangle, and the points $x, \hat{x}, y$ form a similar right triangle. The distance from $c$ to $\hat{c}$ is $c_{n+1}$ (provided $c_{n+1} > 0$), and the distance from $x$ to $\hat{x}$ is $x_{n+1}$. It follows that $x_{n+1}/c_{n+1} = 1 - 1/T$.
Best Answer
No, it's not equivalent at all. You can for example fuse two triangular pyramids together to form a triangular bipyramid. This polyhedron has six equal faces which happens to be triangles instead (they could be equilateral). Since triangles are congruent if their side are the same and we can assign the edge lengths quite freely on such figure we don't need the triangles to be equilateral or even isosceles.
Another counter example is a rhombohedron. Here all the edges are the same, but the faces are rhobi. Given the side length of a rhomb it's determined up to congruence by the angle at one corner. In the same way the rhombohedron is determined by its edge length and the angles at one corner and the angles at the oposite corner will match due to parallellity making all faces congruent.
A more elaborate counter example could be to start with a rhombohedron and then pick two opposite corners and then for each of the other we move them perpendicularily to or from the plane halfway the first corners (perpendicularily to the line connecting them) by equal amount.
Apart from this there are vaugenesses in your proposed definition. The concepts of "equal", "side" is not as precise as one might think. Normally one talks about congruence in geometry (or "similar"). In addition what a "side" is is vaugue: do you mean surface or edge? do you require it to be flat or may it be curved? in that case if it may be a curved surface wheres the border of the surface (you could have a sphere and state that it's made up of six congruent surfaces)?
Having that sorted out one could actually prove that there are basically only these two alternatives (the second is a special case of the third, and the cube is a special case of the second) and counter examples. Such a polyhedron would either have to be assembled the same way as a cube consisting of kite (quadrilateral where each edge has an adjacent edge of the same length) surfaces or assembled like a triangular bipyramid.
The proof is by considering a corner and then rule out the possibility that other than three faces meet there. Then one rules out that they be anything else than triangles or quadrilateral. Last one rules out that the quadrilateral can be anything else than kites (by ruling out anything else than pairs of edges with the same length and that they can't be parallellograms unless they are rhomboids and thereby kites).
After realizing the possibilities one could complete your proposal to apply only to cubes. Going along on the same theme it would be to require twelve equal edges and that all angles are the same.