Let $X_p$ be the set of isometry classes of 4-dimensional positive definite lattices satisfying the property $L^\sharp/L\simeq (\mathbf Z/p)^2$. The set $X_n$ is stable under the involution $\tau : L\mapsto pL^\sharp$.
The question seems to ask whether or not $X_p^\tau$ contains an even lattice. Will Jagy has given evidences for a positive answer. Nevertheless, one might ask whether the same holds when one restricts the question to a given genus (genera are also stable under $\tau$), even or odd.
Further edit Feb. 22 : Interestingly, the case of odd lattices seems to allow the opposite behaviour:
p= 7 A=
[ 7 0 0 0]
[ 0 3 1 -1]
[ 0 1 2 -1]
[ 0 -1 -1 2]
fix= 0 move= 2 proportion= 0.000
p= 23 A=
[23 0 0 0]
[ 0 5 -1 0]
[ 0 -1 3 -1]
[ 0 0 -1 2]
fix= 0 move= 4 proportion= 0.000
p= 31 A=
[31 0 0 0]
[ 0 6 -1 -1]
[ 0 -1 3 0]
[ 0 -1 0 2]
fix= 0 move= 8 proportion= 0.000
p= 47 A=
[47 0 0 0]
[ 0 6 -2 1]
[ 0 -2 5 0]
[ 0 1 0 2]
fix= 0 move= 12 proportion= 0.000
p= 71 A=
[71 0 0 0]
[ 0 7 1 1]
[ 0 1 6 1]
[ 0 1 1 2]
fix= 0 move= 20 proportion= 0.000
p= 79 A=
[79 0 0 0]
[ 0 6 0 1]
[ 0 0 5 -1]
[ 0 1 -1 3]
fix= 0 move= 26 proportion= 0.000
p= 103 A=
[103 0 0 0]
[ 0 18 -1 -1]
[ 0 -1 3 0]
[ 0 -1 0 2]
fix= 0 move= 50 proportion= 0.000
p= 127 A=
[127 0 0 0]
[ 0 10 -2 1]
[ 0 -2 5 -1]
[ 0 1 -1 3]
fix= 0 move= 62 proportion= 0.000
p= 151 A=
[151 0 0 0]
[ 0 11 -1 0]
[ 0 -1 5 1]
[ 0 0 1 3]
fix= 0 move= 78 proportion= 0.000
p= 167 A=
[167 0 0 0]
[ 0 18 -2 -1]
[ 0 -2 5 0]
[ 0 -1 0 2]
fix= 0 move= 92 proportion= 0.000
p= 191 A=
[191 0 0 0]
[ 0 14 0 1]
[ 0 0 5 -1]
[ 0 1 -1 3]
fix= 0 move= 114 proportion= 0.000
p= 199 A=
[199 0 0 0]
[ 0 7 1 1]
[ 0 1 6 0]
[ 0 1 0 5]
fix= 0 move= 134 proportion= 0.000
p= 223 A=
[223 0 0 0]
[ 0 11 3 -3]
[ 0 3 6 -2]
[ 0 -3 -2 5]
fix= 0 move= 182 proportion= 0.000
p= 239 A=
[239 0 0 0]
[ 0 19 2 0]
[ 0 2 7 -1]
[ 0 0 -1 2]
fix= 0 move= 172 proportion= 0.000
Edit Feb. 21 : Here are some heuristics (in each case, there is a quadratic space $V$ on which some lattices $L$ with $q(L)\subset \mathbf Z$ will furnish Gram matrices $A$ such as required in the OP (recall that in that case, the Gram matrix is associated to the bilinear form $x.y=q(x+y)-q(x)-q(y)$, in particular it has even diagonal entries).
Let $p$ be a prime, congruent to $3$ mod $4$. Then there is such a genus on the $\mathbf Q$-quadratic space $[1,1,p,p]$. Here is the number of fixed (resp exchanged) lattices:
p= 3 fix= 1 move= 0 proportion= 1.00
p= 7 fix= 1 move= 0 proportion= 1.00
p= 11 fix= 3 move= 0 proportion= 1.00
p= 19 fix= 3 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 31 fix= 6 move= 0 proportion= 1.00
p= 43 fix= 5 move= 2 proportion= 0.714
p= 47 fix= 15 move= 0 proportion= 1.00
p= 59 fix= 21 move= 0 proportion= 1.00
p= 67 fix= 7 move= 6 proportion= 0.538
p= 71 fix= 28 move= 0 proportion= 1.00
p= 79 fix= 20 move= 2 proportion= 0.909
p= 83 fix= 27 move= 2 proportion= 0.931
p= 103 fix= 25 move= 6 proportion= 0.807
p= 107 fix= 33 move= 6 proportion= 0.846
p= 127 fix= 30 move= 12 proportion= 0.714
p= 131 fix= 65 move= 2 proportion= 0.970
p= 139 fix= 39 move= 12 proportion= 0.765
p= 151 fix= 49 move= 12 proportion= 0.804
p= 163 fix= 15 move= 42 proportion= 0.263
p= 167 fix= 88 move= 6 proportion= 0.937
p= 179 fix= 85 move= 12 proportion= 0.876
p= 191 fix= 117 move= 6 proportion= 0.951
p= 199 fix= 81 move= 20 proportion= 0.802
p= 211 fix= 57 move= 42 proportion= 0.576
p= 223 fix= 70 move= 42 proportion= 0.625
p= 227 fix= 105 move= 30 proportion= 0.777
p= 239 fix= 165 move= 12 proportion= 0.933
p= 251 fix= 161 move= 20 proportion= 0.890
p= 263 fix= 156 move= 30 proportion= 0.839
p= 271 fix= 132 move= 42 proportion= 0.759
p= 283 fix= 75 move= 90 proportion= 0.455
p= 307 fix= 81 move= 110 proportion= 0.424
p= 311 fix= 266 move= 20 proportion= 0.930
p= 331 fix= 87 move= 132 proportion= 0.397
p= 347 fix= 155 move= 110 proportion= 0.585
p= 359 fix= 304 move= 42 proportion= 0.879
p= 367 fix= 144 move= 132 proportion= 0.521
p= 379 fix= 99 move= 182 proportion= 0.353
p= 383 fix= 289 move= 72 proportion= 0.801
p= 419 fix= 333 move= 90 proportion= 0.787
p= 431 fix= 399 move= 72 proportion= 0.847
p= 439 fix= 285 move= 132 proportion= 0.684
p= 443 fix= 195 move= 210 proportion= 0.481
p= 463 fix= 140 move= 272 proportion= 0.340
p= 467 fix= 287 move= 182 proportion= 0.612
p= 479 fix= 525 move= 72 proportion= 0.880
p= 487 fix= 147 move= 306 proportion= 0.325
p= 491 fix= 387 move= 156 proportion= 0.713
p= 499 fix= 129 move= 342 proportion= 0.274
Similarly, in the following cases, there exist such a genus on $A_2\perp p.A_2$:
p= 5 fix= 1 move= 0 proportion= 1.00
p= 11 fix= 3 move= 0 proportion= 1.00
p= 17 fix= 3 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 29 fix= 6 move= 0 proportion= 1.00
p= 41 fix= 10 move= 0 proportion= 1.00
p= 47 fix= 15 move= 0 proportion= 1.00
p= 53 fix= 9 move= 2 proportion= 0.818
p= 59 fix= 21 move= 0 proportion= 1.00
p= 71 fix= 28 move= 0 proportion= 1.00
p= 83 fix= 27 move= 2 proportion= 0.931
p= 89 fix= 27 move= 2 proportion= 0.931
p= 101 fix= 35 move= 2 proportion= 0.946
p= 107 fix= 33 move= 6 proportion= 0.846
p= 113 fix= 22 move= 12 proportion= 0.647
p= 131 fix= 65 move= 2 proportion= 0.970
p= 137 fix= 26 move= 20 proportion= 0.565
p= 149 fix= 49 move= 12 proportion= 0.804
p= 167 fix= 88 move= 6 proportion= 0.937
p= 173 fix= 56 move= 20 proportion= 0.737
p= 179 fix= 85 move= 12 proportion= 0.876
p= 191 fix= 117 move= 6 proportion= 0.951
p= 197 fix= 45 move= 42 proportion= 0.518
p= 227 fix= 105 move= 30 proportion= 0.777
p= 233 fix= 63 move= 56 proportion= 0.529
p= 239 fix= 165 move= 12 proportion= 0.933
p= 251 fix= 161 move= 20 proportion= 0.890
p= 257 fix= 92 move= 56 proportion= 0.622
p= 263 fix= 156 move= 30 proportion= 0.839
p= 269 fix= 132 move= 42 proportion= 0.759
p= 281 fix= 125 move= 56 proportion= 0.690
p= 293 fix= 117 move= 72 proportion= 0.619
Here is a final example : on the $\mathbf Q$-quadratic space $<1,2>\perp p.<1,2>$ (here, the example in the OP, developped in my first answer, appears):
p= 5 fix= 1 move= 0 proportion= 1.00
p= 7 fix= 1 move= 0 proportion= 1.00
p= 13 fix= 1 move= 0 proportion= 1.00
p= 23 fix= 6 move= 0 proportion= 1.00
p= 29 fix= 6 move= 0 proportion= 1.00
p= 31 fix= 6 move= 0 proportion= 1.00
p= 37 fix= 2 move= 2 proportion= 0.500
p= 47 fix= 15 move= 0 proportion= 1.00
p= 53 fix= 9 move= 2 proportion= 0.818
p= 61 fix= 9 move= 2 proportion= 0.818
p= 71 fix= 28 move= 0 proportion= 1.00
p= 79 fix= 20 move= 2 proportion= 0.909
p= 101 fix= 35 move= 2 proportion= 0.946
p= 103 fix= 25 move= 6 proportion= 0.807
p= 109 fix= 15 move= 12 proportion= 0.556
p= 127 fix= 30 move= 12 proportion= 0.714
p= 149 fix= 49 move= 12 proportion= 0.804
p= 151 fix= 49 move= 12 proportion= 0.804
p= 157 fix= 21 move= 30 proportion= 0.412
p= 167 fix= 88 move= 6 proportion= 0.937
p= 173 fix= 56 move= 20 proportion= 0.737
p= 181 fix= 40 move= 30 proportion= 0.571
p= 191 fix= 117 move= 6 proportion= 0.951
p= 197 fix= 45 move= 42 proportion= 0.518
p= 199 fix= 81 move= 20 proportion= 0.802
p= 223 fix= 70 move= 42 proportion= 0.625
p= 229 fix= 50 move= 56 proportion= 0.472
p= 239 fix= 165 move= 12 proportion= 0.933
p= 263 fix= 156 move= 30 proportion= 0.839
p= 269 fix= 132 move= 42 proportion= 0.759
p= 271 fix= 132 move= 42 proportion= 0.759
p= 277 fix= 36 move= 110 proportion= 0.247
p= 293 fix= 117 move= 72 proportion= 0.619
p= 311 fix= 266 move= 20 proportion= 0.930
p= 317 fix= 70 move= 132 proportion= 0.347
p= 349 fix= 105 move= 132 proportion= 0.443
p= 359 fix= 304 move= 42 proportion= 0.879
p= 367 fix= 144 move= 132 proportion= 0.521
p= 373 fix= 80 move= 182 proportion= 0.305
p= 383 fix= 289 move= 72 proportion= 0.801
p= 389 fix= 187 move= 132 proportion= 0.586
p= 397 fix= 51 move= 240 proportion= 0.175
p= 421 fix= 90 move= 240 proportion= 0.273
p= 431 fix= 399 move= 72 proportion= 0.847
p= 439 fix= 285 move= 132 proportion= 0.684
p= 461 fix= 300 move= 156 proportion= 0.658
p= 463 fix= 140 move= 272 proportion= 0.340
p= 479 fix= 525 move= 72 proportion= 0.880
p= 487 fix= 147 move= 306 proportion= 0.325
Studying these repartitions seems to be quite interesting.
Original post :
The answer is no : if you run the following Magma code :
M:=Matrix(Integers(),4,4,[1,0,1,1,0,2,1,2,0,0,5,1,0,0,0,10]);
M:=M+Transpose(M);
L:=LatticeWithGram(M);
H:=GenusRepresentatives(L);
for h in H do
print "h= lattice with Gram", GramMatrix(h);
hd:=DualBasisLattice(h);
MD:=37*LLLGram(GramMatrix(hd));
print "rescaled dual = lattice with Gram", MD;
a,b:=IsIsometric(h,LatticeWithGram(MD));
print "are isometric : ", a;
print " ";
end for;
on the online calculator, you obtain the following result :
h= lattice with Gram
[ 2 0 1 1]
[ 0 4 1 2]
[ 1 1 10 1]
[ 1 2 1 20]
rescaled dual = lattice with Gram
[ 2 0 -1 -1]
[ 0 4 -1 -2]
[-1 -1 10 1]
[-1 -2 1 20]
are isometric : true
h= lattice with Gram
[ 4 -1 2 1]
[-1 4 -1 0]
[ 2 -1 6 -2]
[ 1 0 -2 20]
rescaled dual = lattice with Gram
[ 2 1 -1 0]
[ 1 8 -4 1]
[-1 -4 12 2]
[ 0 1 2 10]
are isometric : false
h= lattice with Gram
[ 4 1 1 1]
[ 1 6 3 1]
[ 1 3 8 -1]
[ 1 1 -1 10]
rescaled dual = lattice with Gram
[ 4 1 -1 -1]
[ 1 6 -3 -1]
[-1 -3 8 -1]
[-1 -1 -1 10]
are isometric : true
h= lattice with Gram
[ 2 1 0 -1]
[ 1 8 -1 -4]
[ 0 -1 10 -2]
[-1 -4 -2 12]
rescaled dual = lattice with Gram
[ 4 1 1 0]
[ 1 4 2 1]
[ 1 2 6 -2]
[ 0 1 -2 20]
are isometric : false